In the right triangle below b = 18 and c = 20, solve for a in simplified radical form.

Answer:

C11=-2

C22=8

hope this will help you more

good luck

Answer:

1. True

2. False

3. True

4. False

Step-by-step explanation:

The coefficient is -1

Answer:

10.2 is the answer not sure

Step-by-step explanation:

[tex]2\pi {r}^{2} + 2\pi rh[/tex]

[tex]2(3.14) {(20)}^{2} + 2(3.14)(20)(11)[/tex]

[tex]6.28(400) + 6.28(220) \\ 2512 + 1381.6 = 3893.6 {mi}^{2} [/tex]

Triangle is always 180 degrees because the sum of all angles in a triangle is 180 degrees. This is known as the "Triangle Angle Sum Theorem." The theorem states that the sum of all angles in a triangle is equal to 180 degrees.

1. A triangle is a three-sided polygon with three angles.

2. The sum of the three angles in any triangle must be 180 degrees.

3. This is known as the Triangle Angle Sum Theorem.

4. The Triangle Angle Sum Theorem states that the sum of the angles in a triangle must equal 180 degrees.

5. It does not matter which angles the triangle has, the sum will always equal 180 degrees.

6. This theorem is based on the fact that the sum of the angles in any triangle is equal to two right angles or 180 degrees.

7. The Triangle Angle Sum Theorem can be proven by drawing a triangle, and then measuring and adding the three angles.

8. The sum of the three angles will always equal 180 degrees.

9. Therefore, triangle is always 180 degrees.

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Answer:

x (blue circle thing) = 46

y (dna) = -63

z (molecules) = 52

Step-by-step explanation:

The solution to the equations is

The value of first x = € 160

The value of second y = € 120

The value of third z = € 140

Equations are mathematical statements with two algebraic expressions flanking the equals (=) sign on either side.

It demonstrates the equality of the relationship between the expressions printed on the left and right sides.

Coefficients, variables, operators, constants, terms, expressions, and the equal to sign are some of the components of an equation. The "=" sign and terms on both sides must always be present when writing an equation.

Given data ,

Let the equation be represented as A

Now , the value of A is

Let the first , second and third be represented as x , y , and z

The total amount to be distributed = € 420

So , x + y + z = € 420   be equation (1)

Now , first has €40 more than the second

So , x = y + 40   be equation (2)

And , third has half of what they have between the first two

So , z = ( 1/2 ) ( x + y )   be equation (3)

Substituting the value of x in the equation (3) , we get

z = ( 1/2 ) ( 2y + 40 )

z = y + 20   be equation (4)

Substituting the value of z and x in equation (1) , we get

( y + 40 ) + y + ( y + 20 ) = 420

On simplifying the equation , we get

3y + 60 = 420

Subtracting 60 on both sides of the equation , we get

3y = 360

Divide by 3 on both sides of the equation , we get

y = € 120

Therefore , the value of x = € 160

The value of z = € 140

Hence , the solution to the equation is € 160 + € 120 + € 140 = € 420

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Answer:

D

Step-by-step explanation:

find area of first rectangle

18 x 6 = 108

find area of second rectangle

20 x 5 = 100

find area of two triangles

7 x 8 =56

56/2 = 28

each triangle has an area of 28

28+28+100+108 = 264ft

Answer:

Helloooooooooo theree

Answer:

don't

Step-by-step explanation: it hard

Answer:

140

Step-by-step explanation:

60+80=?......All you need to do is just add 60 and 80 and get the product

I hope that helped.

Answer:

i) The angular difference in their longitude is approximately 50.65°

ii) The line of longitude where B is located is approximately 30.65°E

Step-by-step explanation:

The given parameters are;

The angle of latitude where the points A and B are located = 60°S

The distance between points A and B = 2,816 km

(i) The angular difference in their longitude, θ, is given as follows;

θ = 2,816/(2 × π × 6,371 × cos(60°)) × 360 ≈ 50.65°

The angular difference in their longitude, θ ≈ 50.65°

(ii) The location of the point A relative to B = Due west

The line of longitude on which A is located = 20°W

∴ The line of longitude where B is located, P = θ - 20°

∴ P = 50.65° - 20°W = 30.65°E

The line of longitude where B is located ≈ 30.65°E.

Given:

Quadrilateral STUV is rotated 270° counterclockwise about the origin.

The coordinates of V were (3, -4).

To find:

The coordinates of V’.

Solution:

If a figure rotated 270° counterclockwise about the origin, then the rule of rotation is:

[tex](x,y)\to (y,-x)[/tex]

The coordinates of V were (3, -4). Using the above rotation rule, we get

[tex]V(3,-4)\to V'(-4,-3)[/tex]

Therefore, the coordinates of V’ are (-4,-3).

Answer:

y=4x+2

Step-by-step explanation:

-8x+2y=4

y=4x+2

−8x+2(4x+2)=4-8x+2(4x+2)=4

y=4x+2

−8x+8x+4=4-8x+8x+4=4

y=4x+2

0+4=40+4=4

y=4x+2

4=4

y=4x+2

Answer:

Image attached

Step-by-step explanation:

Origin looks to be (0, -1), so move each point out by 2

Answer:

5.1

Step-by-step explanation:

so the leaning ladder makes a right angled triangle so we can use pythagoras to calculate the the length (the hypotenuse):

5²+1²=x²

25+1=x²

√26=x

x=5.099...

and now we round to one d.p so it becomes 5.1 (1 d.p) as the hundreths digit is 9 and 9 is rounded up.

Answer:

D)  (6.6, 7.4)

Step-by-step explanation:

Begin by finding the margin of error based on the given information.

[tex]\boxed{\begin{minipage}{7.3cm}\underline{Margin of Error formula}\\\\$\textsf{Margin of Error}= Z \times \dfrac{S}{\sqrt{n}}$\\\\where:\\\phantom{ww} $\bullet$ $Z =$ $Z$ score\\\phantom{ww} $\bullet$ $S =$ Standard Deviation of a population\\\phantom{ww} $\bullet$ $n =$ Sample Size\\\end{minipage}}[/tex]

Given:

The z-score for 95% confidence level is 1.96.

Therefore, to find the margin of error, substitute the values into the formula:

[tex]\implies \textsf{Margin of Error}= 1.96 \times \dfrac{1.3}{\sqrt{50}}[/tex]

[tex]\implies \textsf{Margin of Error}=0.360341...[/tex]

To find a reasonable range for the true mean number of hours a teenage spend on their phone, subtract and add the margin of error to the given mean of 7 hours:

[tex]\begin{aligned}\implies \textsf{Lower bound}&=7-0.360341...\\&=6.63965...\\&=6.6\;\; \sf (1\;d.p.)\end{aligned}[/tex]

[tex]\begin{aligned}\implies \textsf{Upper bound}&=7+0.360341...\\&=7.360341...\\&=7.4\;\; \sf (1\;d.p.)\end{aligned}[/tex]

Therefore, the reasonable range for the true mean is:

-20w-35

I used the distributive property to distribute terms, and then simplified

No solution is when a system of equations has no solution. This means that the equations are inconsistent, and there is no set of values that can make the equations true.

For example, if the system of equations is 2x + 3y = 4 and 2x + 3y = 5, then there is no solution. This is because no matter what values are chosen for x and y, the equations cannot both be true. This means that the system of equations is inconsistent, and there is no solution.

To symbolically represent no solution, you can use the following notation:

2x + 3y = 4, 2x + 3y = 5   No solution

This notation is used to indicate that the equations are inconsistent and that there is no solution.

Complete question: What is the solution to the equation 2x + 3y = 4 and 2x + 3y = 5?

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Answer:

{-1, 5} (two real solutions).

Step-by-step explanation:

On the left side we want x^2, not x?

We need to rewrite x? = 4x - 5 in standard form, that is, as a quadratic with x terms in decreasing powers of x:

x^2 - 4x + 5 = 0

This factors easily to (x - 5)(x + 1) = 0.

We let each factor equal zero (0) separately and solve for x:  

{-1, 5} (two real solutions).

The number of times which the two computers back up at the same time in twenty four hours as required is; 144.

It follows from the task content that the first computer backs up after every five minutes and the second computer backs up every two minutes.

On this note, the least common multiple of both backup times represents when the two computers back up at the same time.

Hence, since the LCM of 5 and 2 is; 10.

It follows that the two computers back up at the same time after every ten minutes.

Ultimately, the number of times for which they backup at the same time in 24 hours; since 60 minutes = 1 hour is;

= ( 24 × 60 ) / 10

= 24 × 6

= 144 times.

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A linear, negative relationship is the type of relationship which is shown in graph of Time vs. Position of Battery Operated Car .

A linear equation is an algebraic expression such as y=mx+b. The slope is m, and the y-intercept is B. A "linear equation in two variables" is the term used to describe the previous sentence, which has variables named y and x. Two different variables can be found in bivariate linear equations. Examples of linear equations include 2x - 3 = 0, 2y = 8, m + 1 = 0, x/2 = 3, x + y = 2, and 3x - y + z = 3. The term "linear" refers to an equation having the form y=mx+b, where m stands for the slope and b for the y-intercept.

given

The area's 46-16 opposed to 14 is different from the area's 18 and T in a different way. The place is there. It isn't linearly getting farther apart. The lee is actually growing as a result.

Option B, the slanting upwards, is what we're going to go with. Since it will be linear, it isn't option C. It is untrue that the speaking has not been constant over the distance. Because it is a D-option, this increase-decrease is untrue. Option B is the best choice. Absolutely.

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Therefore, 60 different combinations can be made wearing one of each type

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Answer:

6

Step-by-step explanation:

6

The basis for the eigenspace corresponding to lambda=5,1,4 are None,[tex]\left[\begin{array}{c}-1 \\\frac{1}{2} \\0\end{array}\right][/tex] and [tex]$\left[\begin{array}{l}2 \\ 1 \\ 1\end{array}\right]$[/tex]

[tex]$$A=\left[\begin{array}{ccc}5 & -12 & 10 \\0 & 7 & -3 \\0 & 6 & -2\end{array}\right]$$[/tex]

Eigenspace corresponding to lambda=5,1,4

The eigenspace E_lambda corresponding to the eigenvalue lambda is the null space of the matrix a [tex]\mathrm{A}-(\lambda) \mathrm{I}"[/tex]

for lambda=5

[tex]$$\mathrm{E}_5=\mathrm{N}(\mathrm{A}-5 \mathrm{I})$$[/tex]

Reducing the matrix A-5I by elementary row operations

[tex]$$\begin{aligned}A-5 I & =\left[\begin{array}{ccc}5-5 & -12 & 10 \\0 & 7-5 & -3 \\0 & 6 & -2-5\end{array}\right] \\& =\left[\begin{array}{ccc}0 & -12 & 10 \\0 & 2 & -3 \\0 & 6 & -7\end{array}\right] \\& \sim\left[\begin{array}{ccc}0 & -12 & 10 \\0 & 1 & -\frac{3}{2} \\0 & 6 & -7\end{array}\right] R_2 \rightarrow \frac{R_2}{2} \\& \sim\left[\begin{array}{ccc}1 & 0 & -8 \\0 & 1 & -\frac{3}{2} \\0 & 6 & -7\end{array}\right] R_1 \rightarrow R_1+2 R_2\end{aligned}$$[/tex]

[tex]\sim\left[\begin{array}{ccc}1 & 0 & -8 \\ 0 & 1 & -\frac{3}{2} \\ 0 & 0 & 2\end{array}\right] R_3 \rightarrow R_3-6 R_2$$\\\sim\left[\begin{array}{ccc}1 & 0 & -8 \\ 0 & 1 & -\frac{3}{2} \\ 0 & 0 & 1\end{array}\right] R_3 \rightarrow \frac{\mathrm{R}_3}{2}$$\\\sim\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & -\frac{3}{2} \\ 0 & 0 & 1\end{array}\right] \mathrm{R}_1 \rightarrow \mathrm{R}_1+8 \mathrm{R}_3$[/tex]

[tex]$\sim\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] R_2 \rightarrow R_2+\frac{2 R_3}{2}$[/tex]

The solutions x of A-5I=0 satisfy x_1=x_2=x_3=0 that is, the null space solves the matrix

[tex]$$\left[\begin{array}{lll}1 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & 1\end{array}\right]\left[\begin{array}{l}x_1 \\x_2 \\x_3\end{array}\right]=\left[\begin{array}{l}0 \\0 \\0\end{array}\right]$$[/tex]

Hence The null space is [tex]\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right] E_5[/tex] has no basis

[tex]$$\begin{aligned}& \text { case: } 2 \\& \text { for } \lambda=1 \\& \mathrm{E}_5=\mathrm{N}(\mathrm{A}-(1) \mathrm{I})\end{aligned}$$[/tex]

we reduce the matrix A-I by elementary row operations as follows.

[tex]$$\begin{aligned}A-1 & =\left[\begin{array}{ccc}5-1 & -12 & 10 \\0 & 7-1 & -3 \\0 & 6 & -2-1\end{array}\right] \\& =\left[\begin{array}{ccc}1 & -3 & \frac{5}{2} \\0 & 6 & -3 \\0 & 6 & -3\end{array}\right] R_1 \rightarrow \frac{R_1}{4} \\& \sim\left[\begin{array}{ccc}1 & -3 & \frac{5}{2} \\0 & 1 & -\frac{1}{2} \\0 & 6 & -3\end{array}\right] R_2 \rightarrow \frac{R_2}{6}\end{aligned}[/tex]

[tex]$$$\sim\left[\begin{array}{ccc}1 & 0 & 1 \\ 0 & 1 & -\frac{1}{2} \\ 0 & 6 & -3\end{array}\right] R_1 \rightarrow R_1+3 R_2$\\$\sim\left[\begin{array}{ccc}1 & 0 & 1 \\ 0 & 1 & -\frac{1}{2} \\ 0 & 0 & 0\end{array}\right] R_3 \rightarrow R_3-6 R_2$[/tex]

Thus, the solutions x of (A-I) X=0 satisfy

[tex]$\left[\begin{array}{ccc}1 & 0 & 1 \\ 0 & 1 & -\frac{1}{2} \\ 0 & 0 & 0\end{array}\right]\left[\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right]=\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]$[/tex]

x_3=t

[tex]$\Rightarrow \mathrm{x}_1=-\mathrm{t}, \mathrm{x}_2=\frac{\mathrm{t}}{2}$[/tex]

[tex]$\vec{x}=\left[\begin{array}{c}-t \\ \frac{t}{2} \\ t\end{array}\right]=\left[\begin{array}{c}-1 \\ \frac{1}{2} \\ 1\end{array}\right] t$[/tex]

The Basis for the nullspace A-I will be: [tex]$\left.\left(\begin{array}{c}-1 \\ \frac{1}{2} \\ 1\end{array}\right]\right)$[/tex]

case:3

lambda=4

[tex]$$\mathrm{E}_5=\mathrm{N}(\mathrm{A}-(4) \mathrm{I})$$[/tex]

we reduce the matrix A-4I by elementary row operations as follows.

[tex]$\begin{aligned} A-4 \mid & =\left[\begin{array}{ccc}5-4 & -12 & 10 \\ 0 & 7-4 & -3 \\ 0 & 6 & -2-4\end{array}\right] \\ & =\left[\begin{array}{ccc}1 & -12 & 10 \\ 0 & 3 & -3 \\ 0 & 6 & -6\end{array}\right] \\ & \sim\left[\begin{array}{ccc}1 & -12 & 10 \\ 0 & 1 & -1 \\ 0 & 6 & -6\end{array}\right] R_2 \rightarrow \frac{R_2}{3}\end{aligned}$[/tex]

[tex]$\begin{aligned} & \sim\left[\begin{array}{ccc}1 & 0 & -2 \\ 0 & 1 & -1 \\ 0 & 6 & -6\end{array}\right] \mathrm{R}_1 \rightarrow \mathrm{R}_1+12 \mathrm{R}_2 \\ & \sim\left[\begin{array}{ccc}1 & 0 & -2 \\ 0 & 1 & -1 \\ 0 & 0 & 0\end{array}\right] \mathrm{R}_3 \rightarrow \mathrm{R}_3-6 \mathrm{R}_2\end{aligned}$[/tex]

Thus, the solutions x of (A-4IX)=0 satisfy

[tex]$$\left[\begin{array}{ccc}1 & 0 & -2 \\0 & 1 & -1 \\0 & 0 & 0\end{array}\right]\left[\begin{array}{l}x_1 \\x_2 \\x_3\end{array}\right]=\left[\begin{array}{l}0 \\0 \\0\end{array}\right]$$[/tex]

x_3=t

[tex]$\Rightarrow \mathrm{x}_1=2 \mathrm{t}, \mathrm{x}_2=\mathrm{t}$[/tex]

[tex]$$\vec{x}=\left[\begin{array}{c}2 t \\t \\t\end{array}\right]=\left[\begin{array}{l}2 \\1 \\1\end{array}\right] t$$[/tex]

The Basis for the nullspace A-4 I will be [tex]\left(\left[\begin{array}{l}2 \\ 1 \\ 1\end{array}\right]\right)[/tex]

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The measures of the angles of the parallelogram are

The measure of ∠A = 72°

The measure of ∠B = 108°

A parallelogram is a simple quadrilateral with two pairs of parallel sides. The opposite or facing sides of a parallelogram are of equal length and the opposite angles of a parallelogram are of equal measure

The four types are parallelograms, squares, rectangles, and rhombuses

Properties of Parallelogram

Opposite sides are parallel

Opposite sides are congruent

Opposite angles are congruent

Same-Side interior angles (consecutive angles) are supplementary

Each diagonal of a parallelogram separates it into two congruent triangles

The diagonals of a parallelogram bisect each other

Given data ,

Let the parallelogram be represented as ABCD

Now , the measure of ∠A = ( 5y - 3 )°

The measure of ∠C = ( 3y + 27 )°

For a parallelogram , opposite angles are congruent

So , measure of ∠A = measure of ∠C

Substituting the values in the equation , we get

( 5y - 3 )° =  ( 3y + 27 )°

On simplifying the equation , we get

( 5y - 3 ) =  ( 3y + 27 )

Subtracting 3y on both sides of the equation , we get

2y - 3 = 27

Adding 3 on both sides of the equation , we get

2y = 30

Divide by 2 on both sides of the equation , we get

y = 15

So , the measure of ∠A = ( 5y - 3 )°

The measure of ∠A = ( 5 ( 15 ) - 3 ) )°

The measure of ∠A = ( 75 - 3 )°

The measure of ∠A = 72°

Now , for a parallelogram , same-side interior angles (consecutive angles) are supplementary

So , the measure of ∠A + the measure of ∠B = 180°

Substituting the values in the equation , we get

72° + the measure of ∠B = 180°

Subtracting 72° on both sides of the equation , we get

The measure of ∠B = 108°

Hence , the measures of angles of the parallelogram are 72° and 108° respectively

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The tasks that Keith can delegate to the assistant are List the account balances from the general ledger in the Trial Balance columns of the end-of-period spreadsheet, Add the Debit and Credit columns of the trial balance and Type the formal financial statements using the data from the Income Statement and Balance Sheet columns of the spreadsheet.

List the account balances from the general ledger in the Trial Balance columns of the end-of-period spreadsheet: Keith should delegate this task to his assistant. This is a relatively straightforward task that does not require a deep understanding of accounting principles. The assistant can be provided with clear instructions on how to pull the account balances from the general ledger and list them in the trial balance columns of the spreadsheet.

Add the Debit and Credit columns of the trial balance: Keith should delegate this task to his assistant. This is a simple arithmetic task that the assistant can perform with clear instructions.

Make the adjusting entries on the spreadsheet: Keith should retain this task. Adjusting entries involve more complex accounting knowledge and require a deeper understanding of accounting principles. It would be difficult for the assistant to complete this task without proper training.

Complete the spreadsheet: Keith should retain this task. This task requires a deep understanding of accounting principles and an understanding of how the various accounts and transactions relate to each other. It would be difficult for the assistant to complete this task without proper training.

Type the formal financial statements using the data from the Income Statement and Balance Sheet columns of the spreadsheet: Keith should delegate this task to his assistant. The assistant can be provided with clear instructions on how to organize the data from the spreadsheet into formal financial statements. It does not require a deep understanding of accounting principles.

Journalize and post the adjusting entries: Keith should retain this task. This task requires a deep understanding of accounting principles and an understanding of how the various accounts and transactions relate to each other. It would be difficult for the assistant to complete this task without proper training.

--The question is incomplete, answering to the question below--

"Keith Martin is the controller for Daniels Printing Service. Keith has been putting in a lot of overtime; therefore, Mr. Daniels has allowed Keith to hire an assistant. Keith's assistant is a bright, high school graduate, but he has never taken an accounting class. Keith is trying to decide which accounting activities could be delegated to his assistant. Keith is willing to give the assistant a few simple instructions on how to complete each task, but he doesn't have the time to teach the assistant to be an accountant.

For each task listed, state whether Keith should continue to do the work or delegate the task to his assistant. Explain each answer.

1. List the account balances from the general ledger in the Trial Balance columns of the work sheet.

2. Add the Debit and Credit columns of the trial balance.

3. Make the adjusting entries on the work sheet.

4. Complete the work sheet.

5. Type the formal financial statements using the data from the Income Statement and Balance Sheet columns of the work sheet.

6. Journalize and post the adjusting entries."

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Answer:

3 3/4 or 3 6/8

Step-by-step explanation:

hope this helps <3