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Y_Kistochka [10]

3 years ago

6

Simplify.

{y }^{ - 3} " align="absmiddle" class="latex-formula">
Options:
3/y

-1/y^3

-3y

1/y^3

1 answer:

Bas_tet [7]3 years ago

4 0

Answer:

Rewrite the expression using the negative exponent rule [text]{b}^{-n } = 1/bn

Step-by-step explanation:

1/y^3

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I need to find X but I don't know how to do it

klio [65]

I think i am doing this right i am not sure. First you need to add up all the sides (100+38+22+15) it would equal 175 then you need to subtract it from 360 and get the missing value (x) then the answer would be 185

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What is the equation of the line that passes through the point (-1, -7) and has a<br> slope of 2?

slava [35]

Answer:

y=2x-5

Step-by-step explanation:

The slope-intercept form of an equation looks like y=mx+b. Where m is the slope and b is the y-intercept. The slope is already given to us, so m is 2.

To find b use the equation $b=y_1-m*x_1$ . So, to find the y-intercept use b=-7-(2*-1). This equals b=-5.

So, plug in the values to get the final equation y=2x-5.

7 0

2 years ago

Read 2 more answers

Help!!! Find the center and radius of the circle with the equation (x-4)²+(y+6)²=16

WITCHER [35]

Answer:

Center: $(4,-6)$

Radius: $4$

Step-by-step explanation:

It is necessary to remember that the Equation of the circle in Center-radius form, is:

$(x -h)^2 + (y-k)^2 = r^2$

Where the center of the circle is at the point (h, k) and the radius is "r".

Then, given the following equation of the circle:

$(x-4)^2+(y+6)^2=16$

We can notice that it is written in Center-radius form. Then we can identify that:

$h=4\\k=-6$

Therefore the center is:

$(4,-6)$

And the radius is:

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5 0

3 years ago

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xz_007 [3.2K]

I believe you multiply by V

3 0

3 years ago

What is the area of a triangle with vertices (3,0) (9,0) (7,6)

Nezavi [6.7K]

Answer:

The area of the triangle is of 21 units of area.

Step-by-step explanation:

The area of a triangle with three vertices $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ is given by the determinant of the following matrix:

$A = \pm 0.5 \left|\begin{array}{ccc}x_1&y_1&1\\x_2&y_2&1\\1_3&y_3&1\end{array}\right|$

In this question:

Vertices (3,0) (9,0) (7,6). So

$A = \pm 0.5 \left|\begin{array}{ccc}3&0&1\\9&0&1\\7&7&1\end{array}\right|$

$A = \pm 0.5(3*0*1+0*1*7+1*9*7-0*7*1-0*9*1-3*1*7)$

$A = \pm 0.5*(63-21)$

$A = \pm 0.5*42 = 21$

The area of the triangle is of 21 units of area.

5 0

3 years ago