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2 years ago

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Im dying and i cant get up

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Alexxx [7]2 years ago

4 0

Are you serious? For reals?

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What is the slope of this equation?<br> y = 3x - 4

Rainbow [258]

Answer:

m=3 is the slope of this equation y=3x-4

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3 years ago

Consider the function f(x)=15x^2+60x-19 <br> Name the vertex for the function

Assoli18 [71]

The vertex for this function would be (-2,-79)

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2 years ago

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If the endpoints of the diameter of a circle are (−8, −6) and (−4, −14), what is the standard form equation of the circle?

kondaur [170]

Equation of the circle is $(x+6)^{2}+(y+10)^{2}=20$ .

Solution:

The endpoints of the diameter of a circle are (–8, –6) and (–4, –14).

Center of the circle = Mid point of the diameter

Mid point formula:

$$P(x, y)=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)$

Here, $x_1=-8, y_1=-6, x_2=-4, y_2=-14$

$$P(x, y) =\left(\frac{-8-4}{2}, \frac{-6-14}{2}\right)$

$$P(x, y) =\left(\frac{-12}{2}, \frac{-20}{2}\right)$

$$P(x, y) =(-6, -10)$

Center of the circle = (–6, –10)

Radius is the distance between center and any endpoint of the diameter.

To calculate the radius using distance formula.

$r=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

Here, $x_1=-6, y_1=-10, x_2=-8, y_2=-6$

$r=\sqrt{\left(-8-(-6)\right)^{2}+\left(-6-(-10)}\right)^{2}}$

$r=\sqrt{(-8+6)^{2}+(-6+10)^{2}}$

$r=\sqrt{(-2)^{2}+(4)^{2}}$

$r=\sqrt{20}$ units

The standard form of the equation of a circle is

$(x-a)^{2}+(y-b)^{2}=r^{2}$ , where (a, b) are center and r is the radius.

Here, center = (–6, –10) and $r=\sqrt{20}$

$(x-(-6))^{2}+(y-(-10))^{2}={(\sqrt{20})} ^{2}$

$(x+6)^{2}+(y+10)^{2}=20$

Equation of the circle is $(x+6)^{2}+(y+10)^{2}=20$ .

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3 years ago

Please help (Proportions)

Yanka [14]

Answer:

1 cup of sugar ..............

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3 years ago

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A store manager is accepting applications for part-time workers. he can hire no more than 14 people. so far,he has hired 9 peopl

Lorico [155]

The manager already hired 9 people. Let $x$ be the number of people he still can hire. Since these people will add to the 9 he already hired, when the hiring campaign will be over he will have hired $9+x$ people. We know that he can't hire more than 14 people, so the number of people hired must be less than or equal to 14:

$9+x \leq 14$

If we subtract 9 from both sides, we have

$x \leq 14-9=5$

so, the manager can hire at most 5 other people

6 0

3 years ago