Let a = x2 + 4Rewrite the following equation in terms of a and set it equal to zero (x2+4)2+32=12x2+48

What is the length of DE (2 points)<br> What is the length of AE (2 points)

Rudiy27

Answer:

DE=21 AE=57

Step-by-step explanation:

the ratio between sides is the same for DE so 17/34=DE/42

solving you get 21. For AE the radio between is 51/34 times 38= 1.5*38=57

3 0

3 years ago

For each algebraic equation, select the property that could be used to solve it: 1. x-3=-5 2. 6x=72 3. x/5=3

kykrilka [37]

X - 3 = -5........addition property....because u add -3 to both sides

6x = 72.....either division property...because u divide both sides by 6.
or multiplication property...because u multiply both sides by 1/6

x/5 = 3.....multiplication property....because u multiply both sides by 5

5 0

4 years ago

Refer to matrix A and identify the matrix element a31.

Blizzard [7]

<h3>Answer: D) 7</h3>

===================================================

Explanation:

a31 means "the value in the third row, first column". That value is 7

As another example, a22 = 3 since 3 is in the second row, second column.

Another example: a12 = -1 because we're in the first row and second column.

This is a 3x2 matrix due to it having 3 rows and 2 columns.

5 0

3 years ago

Hi there can anyone help me

SIZIF [17.4K]

The answer is 39 if rounded it is 40

7 0

4 years ago

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) <em>To find the maximum of a function we always use the derivative of the function and we set it equal zero to find the</em><em> critical points.</em>

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

3 years ago