All categories

3 years ago

Math question: Solve for x 2/3(x - 7) = -2

Mathematics

1 answer:

sasho [114]3 years ago

3 0

2(x-7)/3=-2
3*2(x-7)/2*3=-2*3/2
x-7=-2*3/2
x-7=3*-1
x+(7-7)-7-3
x=7-3
x=4

You might be interested in

What is the measure of the base in the triangle shown below?<br> A. 11<br> B. 45<br> C. 25<br> D. 30

dangina [55]

Answer:

c. 25

Step-by-step explanation:

BOTH SIDES HAVE X SO YOU PUT THEM IN A 1 EQUATION WAY AND SOLVE X:

4X+1=2X+23

4X-2X=23-1

2X=32

X= 32/2=11

THEN YOU GO TO THE BASE AND CHANGE X AND SOLVE THE EQUATION WITH X=11

3x11-8

33-8= 25=BASE

5 0

3 years ago

How can you use a graph to check a factorization?

Ray Of Light [21]

Find where the graph crosses the X axis
ex: graph crosses X axis at x=2 and x=6

factorization should be (x-2) and (x-6) for a quadratic

5 0

3 years ago

Find the value of x and y is (3x_6) (12,y-1)

Alika [10]

Answer:

x=4

y=7

Step-by-step explanation:

12=3x

x=4

6=y-1

y=6+1

y=7

8 0

3 years ago

Read 2 more answers

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

Find the surface area of a cylinder with a base radius of 2 yd and a height of 3 yd.

kherson [118]

Answer:

20 π

Step-by-step explanation:

The surface area of a cylinder is made up of 2 circles and its side.

To find the circles' areas, use the area formula for a circle: A = π $r^2$ .

Since our radius is 2 yd, we can substitute r = 2 into this formula.

A = π $2^2$ = 4π

Now remember, this is only the area of once circle. Multiply 4π by 2 to get the area of both circles: 8π.

Now we need to find the surface area of the side.

If we flatten it out, we can see that the side is actually a curled up rectangle.

The length is the circumference of the circle and the width is the height of the cylinder.

To find the circumference, use the circumference formula for a circle: C = 2πr.

Substitute r = 2 into the formula.

C = 2π2 = 4π

Now that we know that the length is 4π, we can multiply the length by the width to find the area of the cylinder's side.

l = 4π

w = 3

A = lw = 4π * 3 = 12π

Finally, we can add the areas of the circles and the side to get the total surface area.

8π + 12π = 20π

And that is your answer!

Please mark as Brainliest! :)

3 0

2 years ago