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2 years ago

13

Which of the following represents a direct variation?

1 answer:

andrey2020 [161]2 years ago

6 0

Direct variation should be called linear variation.

In this case, the answer is B.

A has the quadratic variation, C has the inverse variation, D has the cubic variation

You might be interested in

((-2^2)(-1^-3))•((-2^-3)(-1^5))^-2

Vitek1552 [10]

Answer:

256

Step-by-step explanation:

A calculator works well for this.

_____

None of the minus signs are subject to the exponents (because they are not in parentheses, as (-1)^5, for example. Since there are an even number of them in the product, their product is +1 and they can be ignored.

1 to any power is still 1, so the factors (1^n) can be ignored.

After you ignore all of the things that can be ignored, your problem simplifies to ...

(2^2)(2^-3)^-2

The rules of exponents applicable to this are ...

(a^b)^c = a^(b·c)

(a^b)(a^c) = a^(b+c)

Then your product simplifies to ...

(2^2)(2^((-3)(-2)) = (2^2)(2^6)

= 2^(2+6)

= 2^8 = 256

3 0

2 years ago

H=4(x+3y)+2 make x the subject

blagie [28]

H = 4(x + 3y) + 2
H = 4x + 12y + 2
H - 12y - 2 = 4x
(H - 12y - 2) / 4 = x or 1/4H - 3y - 1/2 = x

5 0

2 years ago

Read 2 more answers

The table shows list of walking trails in the United States. Latisha walked the KATY trail six days last week. How many miles di

myrzilka [38]

How long is the trail on your table then multiply by 6.

7 0

3 years ago

Fill in the missing portions of the function to rewrite g(x) = 3a^2 − 42a + 135 to reveal the zeros of the function. What are th

mariarad [96]

Answer:

g(x) = 3(x-9)(x-5)

Zeros: x = 9 and x = 5.

Step-by-step explanation:

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$ .

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})$ , given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}$

$\bigtriangleup = b^{2} - 4ac$

In this question:

$g(x) = 3x^{2} - 42a + 135$

So

$a = 3, b = -42, c = 135$

$\bigtriangleup = (-42)^{2} - 4*3*135 = 144$

$x_{1} = \frac{-(-42) + \sqrt{144}}{2*3} = 9$

$x_{2} = \frac{-(42) - \sqrt{144}}{2*3} = 5$

So

g(x) = 3(x-9)(x-5)

Zeros: x = 9 and x = 5.

8 0

3 years ago

Find the values of m and b.<br> M=<br> B=

Stella [2.4K]

Answer:

M= -3

B=6

Step-by-step explanation:

6-0=6

0-2=-2

=3

4 0

2 years ago