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3 years ago

how do I find the area of an equilateral triangle given the perimeter of 60 and the height of 17.3 cm?

1 answer:

3 0

If perimeter is 60 so the side is 60 ÷ 3 = 20

Now we can apply the area formula for triangles;

$A=\frac{1}{2}.b.h\\
\\
A=\frac{1}{2}.20.17.3=10.17.3=173 \ cm^2$

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A rental company charges $15 plus $4 per hour to rent a bicycle. If Margie does not want to spend more than $27 for her rental,

serg [7]

I believe it’s 3 hours

3 0

3 years ago

AC has endpoints<br> A(3,7) and C(6,11).<br> Find AB if B is the<br> midpoint of AC.

faltersainse [42]

Answer:

d=2.5

Step-by-step explanation:

first find the coordinate of B(mid point of AC):A(3,7) C(6,11)

d=√(6-3)²+(11-7)²

d=√3²+4²

d=√9+16=√25=5

since B is the mid point : d/2=5/2=2.5

<h2>Another way :</h2>

B(x1+x2/2 , y1+y2/2) , x1=3 , x2=6, y1=7, y2=11

B(9/2,18/2)

B(9/2,9)

Find AB : the length or distance between 2 points:

d=√(x2-x1)²+(y2-y1)²

d=√(3-9/2)²+(7-9)²

d=√(-3/2)²+(-2)²

d=√1.5²+4

d=√6.25

d=2.5

8 0

3 years ago

If <img src="https://tex.z-dn.net/?f=%20g%28x%29%3D%5Cfrac%7B1%7D%7Bx%7D%20%20" id="TexFormula1" title=" g(x)=\frac{1}{x} " alt

Ksju [112]

Fill in the given values and simplify.

$\dfrac{g(x+h)-g(x)}{h}=\dfrac{\dfrac{1}{x+h}-\dfrac{1}{x}}{h}\\\\=\dfrac{x-(x+h)}{h\cdot x(x+h)}\\\\=\dfrac{-1}{x(x+h)}$

4 0

3 years ago

Im solving variables i need to simplify -5x + 3 (x+5) = -3

victus00 [196]

Answer:

x=9

Step-by-step explanation:

Add similar elements: -5x+3x=-2x

-2x+15=-3

Subtract 15 from both sides

-2x + 15-15=-3-15

Simplify

-2x=-18

Divide both sides by -2

-21/-2 = -18/-2

Simplify

x=9

8 0

3 years ago

Read 2 more answers

Plot the corresponding points for the inverse of the function represented by this table of values. x -4 0 3 7

NNADVOKAT [17]

Answer:

See explanation

Step-by-step explanation:

Given the function f(x) for which

$\begin{array}{ccccc}x&-4&0&3&7\\ \\f(x)&9&5&2&-2\end {array}$

If you want to plot the corresponding points for the inverse of the function f(x), change $f(x)$ into $y$ :

$\begin{array}{ccccc}x&-4&0&3&7\\ \\y&9&5&2&-2\end {array}$

switch x and y

$\begin{array}{ccccc}y&-4&0&3&7\\ \\x&9&5&2&-2\end {array}$

and then change $y$ into $f^{-1}(x).$

You get

$\begin{array}{ccccc}x&9&5&2&-2\\ \\f^{-1}(x)&-4&0&3&7\end {array}$

Now plot these points on the coordinate plane (see attached diagram).

3 0

3 years ago