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3 years ago

15

How can 3 tents be 40 dollars

1 answer:

nasty-shy [4]3 years ago

5 0

I'm assuming this is a math question and all the tents equal the same thing, so to find the price for one tent, we would divide 40/3

Each tent is about $13.33

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

4 0

3 years ago

Please help me because I need to turn this on tommorrow

Trava [24]

YAY I KNOW HOW TO DO THIS OKAY
so 90 degrees minus 49 degrees equals 41 degrees
then 41= (x+3) then you subtract 3
so x=38

7 0

3 years ago

Read 2 more answers

Solve |p + 2| = 10Solve |p + 2| = 10<br><br> {-12}<br> {-8, 8}<br> {-12, 8}

valentinak56 [21]

By definition, we have

$|p+2| = \begin{cases} p+2 &\text{ if } p+2 \geq 0 \\-p-2 &\text{ if } p+2 < 0 \end{cases}$

So, we have to solve two different equations, depending of the possible range for the variable. We have to remember about these ranges when we decide to accept or discard the solutions:

Suppose that $p+2\geq 0 \iff p \geq -2$

In this case, the absolute value doesn't do anything: the equation is

$p+2 = 10 \iff p = 10-2 = 8$

We are supposing $p \geq -2$ , so we can accept this solution.

Now, suppose that $p+2 < 0 \iff p < -2$ . Now the sign of the expression is flipped by the absolute value, and the equation becomes

$-p-2 = 10 \iff -p = 12 \iff p = -12$

Again, the solution is coherent with the assumption, so we can accept this value as well.

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3 years ago

HELP me please with 9,10,11,12

vichka [17]

Answer:

9.) The Outlier is 18

-The outlier is the number that stands out and is different

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2 years ago

Let f(x)=x2+3x−10. Enter the x-intercepts of the quadratic function in the boxes. x = and x =

SIZIF [17.4K]

Answer:

X= 2.

X= -5.

Step-by-step explanation:

Hope it was helpful ;)

3 0

2 years ago