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3 years ago

HELP MEEEEEE!!!!!!!!!!!!!!!!!!!!

Mathematics

1 answer:

SashulF [63]3 years ago

3 0

Answer:

B. ) -2

Step-by-step explanation:

-15 ( x - 2 ) = 25

-15X + 30 = 25

-15X = 25 - 30 15 × 2 = 30

-15X = -5

X = -5 / -15

X = 1 / 3

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Solve for x<br> -1+30x+24=6(5x+4)<br> ???

Tpy6a [65]

Answer:

No solution.

Step-by-step explanation:

Write the expression, combine like terms, and distribute.

-1+30x+24=6(5x+4)

30x+23=30x+24

Get rid of the variable first because they're both 30x (it makes it a lot easier).

30x+23=30x+24

23=24

23 is not equal to 24. When this happens, it means there is no solution, and that's just the answer.

Hope this helps you out!! Have a great day C:

6 0

3 years ago

Teeth rented a truck for one day. There was a base fee of $18.95,and there was an additional charge $.99 for each mile driven. K

Marizza181 [45]

Answer:

138 miles

Step-by-step explanation:

You would have to find the difference between $18.95 and $155.57. When you find the total which is $136.62, you would need to divide it by $0.99. That is what will give you the number of miles that Teeth drove.

6 0

3 years ago

Read 2 more answers

Consider F and C below. F(x, y, z) = y2 sin(z) i + 2xy sin(z) j + xy2 cos(z) k C: r(t) = t2 i + sin(t) j + t k, 0 ≤ t ≤ π (a) Fi

Liono4ka [1.6K]

Answer:

a) $f (x,y,z)= xy^2\sin(z)$

b) $\int_C F \cdot dr =0$

Step-by-step explanation:

Recall that given a function f(x,y,z) then $\nabla f = (\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z})$ . To find f, we will assume it exists and then we will find its form by integration.

First assume that F = $\nabla f$ . This implies that

$\frac{\partial f}{\partial x} = y^2\sin(z)$ if we integrate with respect to x we get that

$f(x,y,z) = xy^2\sin(z) + g(y,z)$ for some function g(y,z). If we take the derivative of this equation with respect to y, we get

$\frac{\partial f}{\partial y} = 2xy\sin(z) + \frac{\partial g}{\partial y}$

This must be equal to the second component of F. Then

$2xy\sin(z) + \frac{\partial g}{\partial y}=2xy\sin(z)$

This implies that $\frac{\partial g}{\partial y}=0$ , which means that g depends on z only. So $f(x,y,z) = xy^2\sin(z) + g(z)$

Taking the derivative with respect to z and making it equal to the third component of F, we get

$xy^2\cos(z)+\frac{dg}{dz} = xy^2\cos(z)$

which implies that $\frac{dg}{dz}=0$ which means that g(z) = K, where K is a constant. So

$f (x,y,z)= xy^2\sin(z)$

b) To evaluate $\int_C F \cdot dr$ we can evaluate it by using f. We can calculate the value of f at the initial and final point of C and the subtract them as follows.