7 is a factor of <br> a 28<br> b 16<br> c 11<br> d 4

HURRY PLSS HELP!!! will give brainliest for right answer!

spayn [35]

9514 1404 393

Answer:

Exponential
Neither

Step-by-step explanation:

The first function has terms with a common ratio of 1/4. It is an exponential function.

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The terms of the second function have neither a common ratio nor a common difference. It is neither exponential nor linear.

4 0

3 years ago

Sorry but plz forgive me for you feeling like I’m using you guys :(

vaieri [72.5K]

Step-by-step explanation:

67yd² is the answer...........

6 0

3 years ago

A machine that makes dimes is adjusted to make them at diameter 1.9 cm. Production records show that when the machine is properl

yanalaym [24]

Answer:

0.0772

Step-by-step explanation:

P(Power) + P(Type II Error) = 1, so P(Type II Error) = 1 – P(Power) = 1 – 0.9228 = 0.0772.

8 0

3 years ago

Help please hllooooo

siniylev [52]

9514 1404 393

Answer:

$81

Step-by-step explanation:

The garden is 4 yards = 12 feet by 3 yards = 9 feet. The area of it is the product of these dimensions, ...

(12 ft)(9 ft) = 108 ft²

The cost of fertilizer for that area will be ...

($0.75/ft²)(108 ft²) = $81.00

They will spend $81.00 to fertilize the garden.

8 0

3 years ago

A particle moving in a planar force field has a position vector x that satisfies x'=Ax. The 2×2 matrix A has eigenvalues 4 and 2

andrey2020 [161]

Answer:

The required position of the particle at time t is: $x(t)=\begin{bmatrix}-7.5e^{4t}+1.5e^{2t}\\2.5e^{4t}-1.5e^{2t}\end{bmatrix}$

Step-by-step explanation:

Consider the provided matrix.

$v_1=\begin{bmatrix}-3\\1 \end{bmatrix}$

$v_2=\begin{bmatrix}-1\\1 \end{bmatrix}$

$\lambda_1=4, \lambda_2=2$

The general solution of the equation $x'=Ax$

$x(t)=c_1v_1e^{\lambda_1t}+c_2v_2e^{\lambda_2t}$

Substitute the respective values we get:

$x(t)=c_1\begin{bmatrix}-3\\1 \end{bmatrix}e^{4t}+c_2\begin{bmatrix}-1\\1 \end{bmatrix}e^{2t}$

$x(t)=\begin{bmatrix}-3c_1e^{4t}-c_2e^{2t}\\c_1e^{4t}+c_2e^{2t} \end{bmatrix}$

Substitute initial condition $x(0)=\begin{bmatrix}-6\\1 \end{bmatrix}$

$\begin{bmatrix}-3c_1-c_2\\c_1+c_2 \end{bmatrix}=\begin{bmatrix}-6\\1 \end{bmatrix}$

Reduce matrix to reduced row echelon form.

$\begin{bmatrix} 1& 0 & \frac{5}{2}\\ 0& 1 & \frac{-3}{2}\end{bmatrix}$

Therefore, $c_1=2.5,c_2=1.5$

Thus, the general solution of the equation $x'=Ax$

$x(t)=2.5\begin{bmatrix}-3\\1\end{bmatrix}e^{4t}-1.5\begin{bmatrix}-1\\1 \end{bmatrix}e^{2t}$

$x(t)=\begin{bmatrix}-7.5e^{4t}+1.5e^{2t}\\2.5e^{4t}-1.5e^{2t}\end{bmatrix}$

The required position of the particle at time t is: $x(t)=\begin{bmatrix}-7.5e^{4t}+1.5e^{2t}\\2.5e^{4t}-1.5e^{2t}\end{bmatrix}$

6 0

3 years ago

7 is a factor of a 28 b 16 c 11 d 4