Each big square below represents one whole.

Please answer this math question for me

klemol [59]

Answer:

angle D is 90⁰

E is 66⁰

and F is 114⁰

7 0

2 years ago

Read 2 more answers

Evaluate C(44,21) use scientific notation, round to 3 decimal places as needed.

STALIN [3.7K]

$C(44,21)=\dfrac{44!}{21!23!}=\dfrac{24\cdot25\cdot\ldots\cdot43\cdot44}{2\cdot3\cdot\ldots\cdot 20\cdot21}=2,012,616,400,080\\
\approx2.013\cdot10^{12}$

6 0

3 years ago

I'd Point A is located at (9,2) on a coordinate plane and point B is located at (-9,10), what is the distance between the two po

Fiesta28 [93]

Draw the triangle and then use Pythagoras' theorem

distance = $\sqrt{18 ^{2} +8 ^{2} }$
= 19.70 (2dp)

4 0

2 years ago

The equation h(t) = -16t² + 80t + 64 represented the height, in feet, of a potato t seconds after it has been launched.

Lyrx [107]

Answer:

Part A) The potato hit the ground at t=5.70 seconds (see the explanation)

Part B) The potato is 40 feet off the ground at the time t=5.28 seconds (see the explanation)

Step-by-step explanation:

we have

$h(t)=-16t^2+80t+64$

where

h(t) is the height of a potato in feet

t is the time in seconds

Part A) Write an equation that can be solved to find when the potato hits the ground. Then solve the equation

we know that

When the potato hit the ground, the value of h(t) must be equal to zero

so

For h(t)=0

$-16t^2+80t+64=0$

Solve the quadratic equation

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$-16t^2+80t+64=0$

so

$a=-16\\b=80\\c=64$

substitute in the formula

$t=\frac{-80\pm\sqrt{80^{2}-4(-16)(64)}} {2(-16)}$

$t=\frac{-80\pm\sqrt{10,496}} {-32}$

$t=\frac{-80+\sqrt{10,496}} {-32}=-0.70$

$t=\frac{-80-\sqrt{10,496}} {-32}=5.70$

therefore

The potato hit the ground at t=5.70 seconds

Part B) Write an equation that can be solved to find when the potato is 40 feet off the ground. Then solve the equation

For h(t)=40 ft

substitute in the quadratic equation

$-16t^2+80t+64=40$

$-16t^2+80t+24=0$

Solve the quadratic equation

we have

$a=-16\\b=80\\c=24$

substitute in the formula

$t=\frac{-80\pm\sqrt{80^{2}-4(-16)(24)}} {2(-16)}$

$t=\frac{-80\pm\sqrt{7,936}} {-32}$

$t=\frac{-80+\sqrt{7,936}} {-32}=-0.28$

$t=\frac{-80-\sqrt{7,936}} {-32}=5.28$

therefore

The potato is 40 feet off the ground at the time t=5.28 seconds

3 0

2 years ago

A 3” x 5” photo is enlarged so that the length of the new photo is 7 inches. Find the width of the new photo.. a. 4.20 inches. c

Andrej [43]

As far as the old dimensions of the photo are concerned,
Width of the photo = 3"
Length of the photo = 5"
New length of the photo = 7 inches
Let us assume the new width of the photo = x inches
Then
3/5 = x/7
5x = 7 * 3
5x = 21
x = 21/5
= 4.20 inches
So the correct option among all the options that are given in the question is the first option. I hope that the answer has come to your help.

3 0

2 years ago