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vovikov84 [41]
2 years ago
6

A car travels 0.75 miles per minute. Explain how you could use proportional reasoning to find how far the car travels in one hou

r
Mathematics
1 answer:
Zina [86]2 years ago
4 0
Notice the proportions below.  Keep in mind that 1 hour is 60 minutes.

\bf \begin{array}{ccll}
miles&minutes\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\
0.75&1\\
x&60
\end{array}\implies \cfrac{0.75}{x}=\cfrac{1}{60}\implies \cfrac{0.75\cdot 60}{1}=x\implies 45=x
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Simplify the expression.<br> 2x³y +7x³y-xy³
jek_recluse [69]

Answer:

xy*(2x^2 + 7x^2 - y^2)

Step-by-step explanation:

You can get xy out of the expression since all have xy

7 0
2 years ago
Name a real life situation in which you would need money to divide ​
oee [108]

Answer:

You and 9 friends all made a bake sale and profited 100 dollars. How much money should you and each friend get.

ANSWER 100/ 9 friends + you

= 100/10

= $10

Step-by-step explanation:

5 0
3 years ago
f a rectangle has a perimeter of 70, a length of x and a width of x - 9, find the value of the length of the rectangle.
trasher [3.6K]

( x + (x  -  9))  \times 2 = 70 \\( 2x - 9) \times2= 70 \\ 4x - 18 = 70 \\ 4x = 70 + 18 = 88
x =  \frac{88}{4}  = 22
the length is 22


and the width is
22-9=17





good luck
3 0
3 years ago
Please help!!! ASAP!
bulgar [2K]

Answer:

so what we have to do first is multiply 9*10

which is 90 now we divide by two to get 45

now we multiply 45 by 20 to get

900

Hope This Helps!!!

6 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
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