SOMEONE PLS HELP ME!!!

Answer:

A

Step-by-step explanation:

using pythagoras theorem

h²=p²+b²

Step-by-step explanation:

smallest value 30° largest value 150°.

Answer:

30, 150

Step-by-step explanation:

sin θ = 1 / 2

sin 30 = 1 / 2

θ = 30

So,

the smallest value of θ is 30.

sin ( 180 - 30 ) = sin 150 = 1 / 2

So,

the larger value of θ = 150.

Answer:

64 by 16 feet

Answer:

left 2 units, up 3 units

Step-by-step explanation:

(x-2, y+3)

-2 means 2 to the left

+3 means 3 up

So it would be left 2 units, up 3 units.

Answer:

Step-by-step explanation:

Rule:

If  x+b   right in b units

if  x-b    left   in b untits

if  y+b    up   in  b units

if  y-b  down in  b  units

You don 't have to use this rule if you know that on the Ox axis the line goes horizontally (the abscissa axis); And on the Oy axis vertically (the ordinate axis)

Then :

x-2    left in 2 units

y+3   up  in  3 units

If v(t) and h(t) denote the ball's velocity and height at time t, respectively, then we have h(0) = 1.2 m and v(0) = 58 m/s.

The ball is in free fall after being tossed, so its acceleration is constant at any time t :

[tex]a(t) = -9.8 \dfrac{\rm m}{\mathrm s^2}[/tex]

By the fundamental theorem of calculus, the velocity function is given by

[tex]\displaystyle v(t) = v(0) + \int_0^t a(u) \, du[/tex]

Compute the integral:

[tex]\displaystyle v(t) = 58 \frac{\rm m}{\rm s} - \left(9.8 \frac{\rm m}{\mathrm s^2}\right) \int_0^t du[/tex]

[tex]\displaystyle v(t) = 58 \frac{\rm m}{\rm s} - \left(9.8 \frac{\rm m}{\mathrm s^2}\right) (t - 0)[/tex]

[tex]\displaystyle \boxed{v(t) = 58 \frac{\rm m}{\rm s} - \left(9.8 \frac{\rm m}{\mathrm s^2}\right) t}[/tex]

Similarly, the height function is given by

[tex]\displaystyle h(t) = h(0) + \int_0^t v(u) \, du[/tex]

[tex]\displaystyle h(t) = 1.2 \, \mathrm m + \int_0^t \left(58 \frac{\rm m}{\rm s} - \left(9.8 \frac{\rm m}{\mathrm s^2}\right) t\right) \, du[/tex]

[tex]\displaystyle h(t) = 1.2 \, \mathrm m + \left(58 \frac{\rm m}{\rm s}\right) (t - 0) - \frac12 \left(9.8 \frac{\rm m}{\mathrm s^2}\right) (t^2 - 0^2)[/tex]

[tex]\displaystyle \boxed{h(t) = 1.2 \, \mathrm m + \left(58 \frac{\rm m}{\rm s}\right) t - \frac12 \left(9.8 \frac{\rm m}{\mathrm s^2}\right) t^2}[/tex]

Answer:

V(t) = 58-9.8t

h(t)=1.2+58t-4.9[tex]t^{2}[/tex]

Step-by-step explanation:

Using the standard formula

a=V f-Vi/t --rearrange

at=V f-Vi

V f=V i + a t

therefore

V(t)=58-9.8t

same with height

H= Hi+Vit+1/2at^2

therefore

h(t)=1.2+58t-4.9t^2

Answer:

1 ream = (3x - 4) pesos

Price of n reams = (6x^4-17x^3+24x^2 34x+24) pesos

Solution:(pic1)

number of reams 3x-4

By using synthetic division,

3x-4=0

3x=4

x=4/3

6 -17 24 -34 24 | 4/3

8-12 16-24

6 -9 12-18 0

Final equation is:(pic2)

Answer:

Step-by-step explanation:

The equation of this line is:

y = -5/2 x + 2.

Answer:

1. 2/5, 2. -1. 3. 5/2, 4. 1, 5. 1/5, 6. -4/3, 7. 1/3, 8. -1/10

Step-by-step explanation:

1-3: the formula for slope is (y2 - y1)/(x2 - x1) for points (x1, y1) and (x2, y2)

1. we have points (-2, -1) and (3,1) ,plug these into the formula to get (1-(-1))/(3-(-2)) = (1+1)/(3+2) = 2/5 so the slope is 2/5

2. we have points (-3, 2) and (2, -3): (-3-2)/(2-(-3)) = -5/5 = -1

3. (-1, -2) (1,3): (3-(-2))/(1-(-1)) = 5/2

4-6: you can use the same formula

4: (1,0) (3,2): (2-0)/(3-1) = 2/2 = 1

5. (8,0) ( 13, 1) (1-0)/(13-8) = 1/5

6. (2, -1) ( 5, -5) (-5+1)/(5-2) = -4/3

7-8: slope= rise/run

7. rise=8, run=24, 8/24=1/3

8. rise=-100, run=1000, slope= -100/1000 = -1/10

[tex]\\ \sf\longmapsto XY=\sqrt{x^2+y^2}[/tex]

[tex]\\ \sf\longmapsto XY=\sqrt{a^2+2^2}[/tex]

[tex]\\ \sf\longmapsto XY=\sqrt{a^2+4}[/tex]

[tex]\\ \sf\longmapsto XY=2+a[/tex]

Answer:

c. 60 mins

Step-by-step explanation:

12 * 5 = 60

15 * 4 = 60

Here

Show sum of interior angles is 180°

[tex]\\ \sf\longmapsto <NBA+<NAB=180[/tex]

[tex]\\ \sf\longmapsto <NAB+103=180[/tex]

[tex]\\ \sf\longmapsto <NAB=180-103[/tex]

[tex]\\ \sf\longmapsto <NAB=77°[/tex]

Vertical angles are supplementary

We need m<NAB

Answer:

0.55%

Step-by-step explanation:

230÷100=2.30%

175÷100=1.75%

2.30-1.75=0.55%

I think ‍♂️

Answer:

-7

Step-by-step explanation:

there are 16 increments in total, and half are labeled

so the answer is -7 because if the increments are 1 and it is left of zero, so do 7*-1 because it is to the left

Answer:

-7

Step-by-step explanation:

Answer:

what is the radius of 2

Step-by-step explanation:

Answer:

you're wrong

Step-by-step explanation:

should have been "Why are all the answers on this website saying my work is wrong?"

Answer:

1.22

Step-by-step explanation:

Answer:

57p

Step-by-step explanation:

so 1 pound is just 1.00 so

1.00-0.43=0.57

so terry will get 57p change

Answer:

9/100

Step-by-step explanation:

9×100 = 9/100 I think

Answer:

161 Remainder 4

Step-by-step explanation:

trust me on this

Answer:

_

161.2 BUT if with remainder it is 161 with a remainder of 4

Step-by-step explanation:

By the way the 2 is a repeating number so make sure to add that line at the top.

Answer:

b)

Step-by-step explanation:

Answer:

look at the picture i have sent

Answer:

see below

Step-by-step explanation:

just reduce to 3 digits before the decimal place and count how many are behind the decimal for what power the 10 is, not including numbers that were already behind the decimal

a. 107.92*10^2

b. 376.932*10^0

c. 998.461*10^3

Answer:

When rounding significant figures the standard rules of rounding numbers apply, except that non-significant digits to the left of the decimal are replaced with zeros.

Step-by-step explanation:

(a) 10792 ≈ 10800

(b) 376.932 ≈ 377

(c) 998461 ≈ 998000

Answer:

5:3=8

3/8 x 120

3/1 x 15

3 x 15=45

Step-by-step explanation:

hope this helps

sorry

Answer:

Should be

Step-by-step explanation:

Answer: Yes

Step-by-step explanation:

Using the formula a^2+b^2=c^2, take the square of 6 (6x6=36) and 8 (8x8=64) and add them together to get the square of 10 (10x10=100) 64+36 does equal 100. Remember that the biggest number is c and the two sides a and b have to add to get c.

Answer:

Angle bisectors, Perpendicular bisectors, and Circumcenters