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Rachel's beginning balance one year ago was $800.

Compound interest is applicable when there will be a change in the principle amount after the given time period.

For example, if you give anyone $500 at the rate of 10% annually then $500 is your principle amount. After 1 year the interest will be $50 and hence principle amount will become $550 now for the next year the interest will be $550, not $500.

Given that,

The current balance is $836 which is the result of compound interest of 4.5 % annually.

The original amount was a priciple amount P on which interest was applied and the final amount $836 made.

Compound interest formula;

A = P[ 1 + RT/100]^T

Where A is the final amount and T is the time period.

Here,

A = 836 , R = 4.5% and T = 1 year.

So,

836 = P[1 + 4.5/100]^1

P = 836/1.045 = $800

Hence "Rachel's beginning balance one year ago was $800".

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Answer:

24 books

Step-by-step explanation:

Books are valued at $10/3 books.  18 books would have a value of (18books)*($10/3books) = $60, and so on.

X books would have a value of X($10/3)=$80

X = ($80)*(3/$10)

X = 24 books

[tex]\text{L.H.S}\\\\=\sin^2 \theta \tan \theta + \cos^2 \theta \cot \theta+2\sin \theta \cos \theta \\\\\\=\sin^2 \theta \cdot \dfrac{\sin \theta}{\cos \theta} + \cos^2 \theta \cdot \dfrac{\cos \theta}{\sin \theta} + 2\sin \theta \cos \theta\\\\\\=\dfrac{\sin^3 \theta}{\cos \theta}+ \dfrac{\cos^3 \theta}{\sin \theta}+ 2\sin \theta \cos \theta\\\\\\=\dfrac{\sin^4 \theta + \cos^4 \theta}{\cos \theta \cdot \sin \theta}+ 2\sin \theta \cos \theta\\\\[/tex]

[tex]=\dfrac{(\sin^2 \theta)^2 + (\cos^2 \theta)^2 }{\cos \theta \sin \theta}+ 2\sin \theta \cos \theta\\\\\\=\dfrac{(\sin^2 \theta + \cos^2 \theta)^2-2\sin^2 \theta \cos^2 \theta}{\cos \theta \sin \theta}+ 2\sin \theta \cos \theta\\\\\\=\dfrac{1-2 \sin^2 \theta \cos^2 \theta}{\cos \theta \sin \theta}+ 2\ sin \theta \cos \theta\\\\\\=\dfrac{1-2\sin^2 \theta \cos^2 \theta+2\sin^2 \theta \cos^2 \theta}{\cos \theta \sin \theta}\\\\=\dfrac 1{\cos \theta \sin \theta}\\\\[/tex]

[tex]=\dfrac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta}\\\\\\=\dfrac{\sin^2 \theta}{\cos \theta \sin \theta}+\dfrac{\cos^2 \theta}{\cos \theta \sin \theta}\\\\\\=\dfrac{\sin \theta}{\cos \theta}+\dfrac{\cos \theta}{\sin \theta}\\\\\\=\tan \theta + \cot \theta\\\\=\text{R.H.S}\\\\\text{Proved.}[/tex]

Answer: 36

Step-by-step explanation: Ratio 27:36 is also 3:4

Answer:

36 games

Step-by-step explanation:

since the ratio is 3:4 we must get what the original is.

27/3=9

27+9=36

27:36

Answer:

A cube is a rectangular prism where all its sides are the same. The formula to find the surface area of a rectangular prism is A = 2wl + 2lh + 2hw, where w is the width, the l is the length, and the h is the height.

The surface area formula for a cube is 6l²

A cube has the following features:

The area of a face of a cube is:

Area = l²

Where l represents the side length

The surface area formula for a cube is then calculated using:

Surface area = 6 * Area of a face

So, we have:

Surface area = 6 * l²

Surface area = 6l²

Hence, the surface area formula for a cube is 6l²

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Answer:

B) formation of a precipitate
E) change in color

Step-by-step explanation:

just did the assignment on e.d.g.e

Answer:

B and E is correct

Step-by-step explanation:

Answer:

A on edge 2021

Step-by-step explanation:

I got it correct.

Answer:

All answers are bold and links are real photos  

BTW THE PHOTOS ARE IN ORDER OF THE ANSWERS

Brainlyist??

Step-by-step explanation:

Task 1

Part A

Tangent Line

Part B

PHOTO WITH LARGE CIRCLE AND LINE OUT SIDE BUT CONNECTED

Part C

2034.7

PHOTO WITH EQUATION

We calculate the point from the center of the earth to the satellite and because we have the radius we can subtract it getting the distance from the satellite to the earth.

Task 2

Gears

Part A

Approximately 10.47 inches

Part B

300°

Part C

Approximately 125.66 inches

Part D

The distance traveled increases by approximately 62.83 inches

Task 3

Circle Theorems

Part A

PHOTO WITH LINE THROUGH CIRCLE AND ONE 90 DEGREE ANGLE

Part B

Prove OC is perpendicular to AB

We know that ΔAEC≅AED by SSS criteria (SSS criteria is when all three sides of a triangle is equal making them congruent)

To prove this that they apply to the SSS criteria: m∠ADO=m∠BDO m∠ADO=90°=m∠BDO

Part C

A radius is perpendicular to a chord when the chord is equally divided in half by the radius.

PHOTO OF LINE THROUGH CIRCLE AND LINE DISTANCES

Part D

Prove OC is perpendicular to AB

AO=BO because AO and BO are both radii

OD≅OD by reflexive theorem

ΔADO≅ΔBDO because they are both right triangle with the hypotenuse and a side

AB≅BC means that OC is perpendicular to AB

Disclaimer: The images to part A and part C is the same.

Two theorems we are proving are:

Part A: The image is attached

Part B:

Let there be a circle with the center at O. Let AC be a chord. Let OE be a radius of the chord such that OE bisects AC at B, that is, AB = BC.

To prove: A line drawn from the center of the circle to bisect a chord is perpendicular to it.

Construction: We join OA and OB.

Proof:

In ΔOAB and ΔOCB,

OA = OC (Radius of the same circle)

AB = BC (Given, at B, the chord is bisected)

OB = OB (common side)

∴ ΔOAB ≅ ΔOCB [SSS axiom]

∴ ∠OBA = ∠OBC [Common parts of Congruent triangles]

Let ∠OBA = ∠OBC = x.

Since AC is a straight line,

∠OBA + ∠OBC = 180° [Linear pair]

or, x + x = 180°

or, 2x = 180°

or, x = 90°.

∴ ∠OBA = ∠OBC = 90°.

So, we can say that OB ⊥ AC, hence the theorem is proved.

Part C: The image is attached

Part D:

Let there be a circle with the center at O. Let AC be a chord. Let OE be a radius of the chord such that OE is perpendicular to AC at B, that is, AB ⊥ BC.

To prove: A line drawn from the center of the circle perpendicular to the chord bisects it.

Construction: We join OA and OB.

Proof:

In ΔOAB and ΔOCB,

OA = OC (Radius of the same circle)

∠OBA = ∠OBC = 90° (∵ OB ⊥ AC)

OB = OB (common side)

∴ ΔOAB ≅ ΔOCB [RHS axiom]

∴ AB = AC [Common parts of Congruent triangles]

So, we can say that OE perpendicular to the chord AC bisects it, hence the theorem is proved.

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Generally, you'd see the equation of a circle organized in the following format:

[tex](x-h)^2+(y-k)^2=r^2[/tex]

To determine the equation given the center and the radius:

We're given:

Plug the radius and center into the equation as r and (h,k):

[tex](x-(-1))^2+(y-(-8))^2=99^2\\(x+1)^2+(y+8)^2=9801[/tex]

[tex](x+1)^2+(y+8)^2=9801[/tex]

yes it is uses in1979 there ie big tsunami

Answer:

the American Embassy in Tehran was overrun and its employees taken captive.

Step-by-step explanation:

Answer:

Pretty sure it's "only i"

Step-by-step explanation:

A constant rate of change means that something changes by the same amount during equal intervals.

$30 is the same amount whereas 10% could vary.

Hope this helps.

7, 7, 9, 27, 27, are the numbers that we can fill these boxes starting from left.

A prime number is not a composite number, prime numbers are only divisible by the number itself and one.

An example of a prime number is 2 which is only divisible by 2 and 1 it is also the only even prime number.

The only prime number is 7.

The intersection of prime and odd numbers is 7.

The only odd number is 9.

The intersection of odd and cube numbers is 27.

The only cube number is 27.

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Answer:

y = 3x -2

Step-by-step explanation:

General equation of a line in the slope intercept form is

y = mx + b ,

where m is the slope and b is the y-intercept ( where intercepts the y-axis)

y = 3x + b ,

because is given slope is m = 3

For point (x = -2, y = -8) we have

-8 = 3 (-2) + b

-8 = -6 + b

-8 + 6 = b

-2 = b

The equation of the line that passes through the point (-2,-8) and has a slope of 3 is

y = 3x -2

Answer:

(1,12) ; (2,9) ; (3,6) ; (4,3) ; (5,0)

Step-by-step explanation:

You need to plug the x values into the equation.

Your equation: y=15-3x

Example:

Plug in 1

y=15-3

y=12

Plug in 2:

y=15-3(2)

y=15-6

y=9

Plug in 3:

y=15-3(3)

y=15-9

y=6

Plug in 4:

y=15-3(4)

y=15-12

y=3

Plug in 5:

y= 15-3(5)

y=15-15

y=0

The total of Melinda's expenses for the week is $93.04 option (c) $91.94 is correct.

It is defined as the money spends on the utility, the amount of money is required to buy something in other words it is the outflow of money from the solely earners income or the money incurred by any organization.

Melinda kept a record of her spending for the week as follows:

The amount she spent money to put gas in the car = $25

The amount to buy a new shirt = $22.38

The amount to play putt-putt golf with some friends = $6.95

The amount to buy groceries = $38.71

To calculate the total expenses for the week, we must add all the expenses:

= $25+$22.38+$6.95+38.71

= $93.04

Thus, the total of Melinda's expenses for the week is $93.04 option (c) $91.94 is correct.

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The difference between the polynomial is [tex]\rm 9x^3-5x^2-6x+10[/tex].

The subtract means to take away from a group or a number of things.

Lorne subtracted 6x3 – 2x + 3 from –3x3 + 5x2 + 4x – 7.

The difference between the expression is;

[tex]\rm ( 6x^3 - 2x + 3 )- (-3x^3 + 5x^2 + 4x - 7)[/tex]

[tex]\rm 6x^3 - 2x + 3 +3x^3 -5x^2 -4x +7\\\\9x^3-5x^2-6x+10[/tex]

Hence, the difference between the polynomial is [tex]\rm 9x^3-5x^2-6x+10[/tex].

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[tex]\\ \rm\dashrightarrow x^{\frac{2}{7}}[/tex]

[tex]\\ \rm\dashrightarrow x^{2(1/7)}[/tex]

[tex]\\ \rm\dashrightarrow \sqrt[7]{x^2}[/tex]

And

[tex]\\ \rm\dashrightarrow \sqrt[3]{2y}[/tex]

[tex]\\ \rm\dashrightarrow 2y^{\frac{1}{3}}[/tex]

Answer:

see below

Step-by-step explanation:

Problem-1:

[tex] {x}^{ \frac{2}{7} } [/tex]

remember that,

[tex]\displaystyle {x}^{ \frac{a}{b} } = \sqrt[b]{ {x}^{a} } [/tex]

here,

hence,

[tex] {x}^{ \frac{2}{7} } = \boxed{\sqrt[7]{ {x}^{2} } }[/tex]

Problem-2:

recall that,

[tex] \sqrt[n]{x} = {x}^{ \frac{1}{n} } [/tex]

therefore,

[tex] \sqrt[3]{2y} = (2y {)}^{ \frac{1}{3} } \implies \boxed{{2 }^{ \frac{1}{3} } {y}^{ \frac{1}{3} } }[/tex]

All polynomials are continuous, so the limit can be evaluated by direct substitution:

[tex]\displaystyle \lim_{x\to-3} (x^3+3x^2+4x-12) = (-3)^3+3(-3)^2+4(-3)-12 = \boxed{-24}[/tex]

The Limit of (x cubed + 3 x squared + 4 x minus 12) as x approaches negative 3 is -24.

Polynomials are those algebraic expressions that consist of variables, coefficients, and constants. The standard form of polynomials has mathematical operations such as addition, subtraction, and multiplication.

All polynomials are continuous, so the limit can be evaluated by direct substitution:

[tex]\lim_{x \to {-3}} x^3 + 3 x ^2+ 4 x - 12\\\\= (-3)^3 + 3 (-3)^2+ 4 (-3)- 12\\\\= -24[/tex]

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Answer:

Yes, I agree with Noah because the first lake has a smaller mean and MAD than the second

Step-by-step explanation:

Heyo!

SaddySokka is here to help!!

Let's do this step-by-step explanation!

[tex]=\frac{5^-^6}{5^3}[/tex]

[tex]=5^-^6^-^3[/tex]

[tex]=5^-^9[/tex]

[tex]=\frac{1}{5^9}[/tex]

Answer:

A. [tex]\frac{1}{5^9}[/tex]

Hopefully, this helps you!!

Have a great day!!

SaddySokka~

The measure of PQ if the triangle LMN is similar to triangle OPQ is 69.4

Similar triangles are triangles that have equal sides and angles. The ratio of the similar sides is a constant.

According to the triangle given:

PQ/17 = OQ/LM
PQ/17 = 49/12
12PQ = 833
PQ = 69.4

Hence the measure of PQ if the triangle LMN is similar to triangle OPQ is 69.4

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Answer:

Step-by-step explanation: y axis is 6units up (y=+6)

X axis is 2units to the right. (×=+2)

The lower quartile is 43.

All the given equations have 2 real solutions except for equation 4 which has only one real root.

A quadratic equation is the second-order degree algebraic expression in a variable. the standard form of this expression is  ax² + bx + c = 0 where a. b are coefficients and x is the variable and c is a constant.

The given quadratic equation

[tex]y = 12x^2 - 9x+ 4[/tex]

where, a = 12, b = -9 and c = 4

x = [tex]-b + \sqrt{b^2 - 4ac} /2a[/tex]

x = [tex]-(-9)+ \sqrt{(-9)^2 - 4(12)(4)} /2(12)[/tex]

x = 1.063 or -0.313

Second Equation

[tex]10x +y = -x^2 +2 \\y = x^{2} +10x - 2[/tex]

where, a = 1, b = 10, c = -2

substitute the values into the equation

x = [tex]-b + \sqrt{b^2 - 4ac} /2a[/tex]

[tex]-10 + \sqrt{10^2 - 4(1)(-2)} /2(1)[/tex]

x = 0.195 or -10.195

Third Equation

[tex]4y - 7 = 5x^2 - x +2 +3y[/tex]

[tex]y = 5x^{2} -x -5[/tex]

where, a = 5, b = -1, c = - 5

x = [tex]-b + \sqrt{b^2 - 4ac} /2a[/tex]

[tex]-(-1) + \sqrt{(-1)^2 - 4(5)(-5)} /2(5)[/tex]

x = 0.905 or -0.55

Fourth Equation

[tex]y = (-x +4)^2 \\y = x^{2} -8x +16[/tex]

where, a = 1, b = -8, c = 16

x = [tex]-b + \sqrt{b^2 - 4ac} /2a[/tex]

x = [tex]-(-8) + \sqrt{(-8)^2 - 4(1)(16)} /2(1)[/tex]

x = 4

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Answer:

y = 12x2 - 9x + 4

✔ no

real solution(s)

4y - 7 = 5x2 - x + 2 + 3y

✔ no

real solution(s)

10x + y = -x2 + 2

✔ two

real solution(s)

y = (-x + 4)2

✔ one

real solution(s)

Step-by-step explanation:

hope this helps:)

Answer:

1. Growth

2. Growth

3. Decay