All categories

3 years ago

Ms. Reynolds is asking students to find

Mathematics

2 answers:

Gemiola [76]3 years ago

7 0

Peter found the correct expression, the width and length of the base are 5 and 14 centimeters.
Thelma found the measurements for the surface on the side of the rectangular pyramid, not the base.

kotegsom [21]3 years ago

7 0

Peter had gotten the answer thelma is the one who got the question wrong

You might be interested in

PLeaseeeee Helppp Me

kow [346]

$y = mx +b\\\\\text{In this case,}~ m = -1~ \text{and}~ b = -8\\\\\\y = (-1)x +(-8) = -x -8$

7 0

3 years ago

Place in order from least to greatest -3, 4, |-5|, 9, -7 Please help :c

natulia [17]

$|-5|=5$

so

$-7,-3,4,|-5|,9$

3 0

3 years ago

What is the answer 7,8,9,10

stepan [7]

7) 116.6666666≈116.67
8) 138.6666666≈138.67
9) 42.1428571
10) 64.6666666≈64.67

3 0

4 years ago

Read 2 more answers

Simplify. (2x5+x3−3x4−x+8)−(7x5+2x4−12+6x−5x3+x2)

Ganezh [65]

Answer:

=−5x5−5x4+6x3−x2−7x+20

Hoped I Helped Have A Great Day

7 0

3 years ago

Read 2 more answers

The height (feet) of an object moving vertically is given by s= (-16t^2)+(208t)+(156), where t is in seconds.

OverLord2011 [107]

Answer:

The object's velocity at t = 7 is -16 ft/s

The maximum height is 832 ft and it occurs when $t=\frac{13}{2}$

Step-by-step explanation:

From the information given, the equation of motion of the object is

$s(t)= -16t^2+208t+156$

For any equation of motion s(t), the instantaneous velocity at time t is given by

$v(t)=\frac{ds}{dt}$

(a) To find the object's velocity at t = 7, you must:

$v(t)=\frac{d}{dt}(-16t^2+208t+156)\\\\v(t)=-\frac{d}{dt}\left(16t^2\right)+\frac{d}{dt}\left(208t\right)+\frac{d}{dt}\left(156\right)\\\\v(t)=-32t+208+0\\\\v(t)=-32t+208$

Next, we evaluate when t = 7

$v(7)=-32(7)+208=-224+208\\\\v(7)=-16$

The object's velocity at t = 7 is -16 ft/s

(b) To find the maximum of a function we always use the derivative of the function and we set it equal zero to find the critical points.

To find the maximum height and when it occurs, we set the velocity function equal to 0 and solve for t.

$-32t+208=0\\\\-32t+208-208=0-208\\\\-32t=-208\\\\\frac{-32t}{-32}=\frac{-208}{-32}\\\\t=\frac{13}{2}$

Next, we substitute this value into the equation of motion to find the maximum height

$s(\frac{13}{2})= -16(\frac{13}{2})^2+208(\frac{13}{2})+156\\\\s(\frac{13}{2})=-676+1352+156)=832$

The maximum height is 832 ft and it occurs when $t=\frac{13}{2}$

6 0

4 years ago