$29 + $4.50x = $42.50
$29-29 + $4.50x = $42.50-$29
$4.50x = $13.50
$4.50x/$4.50 = $13.50/$4.50
x = 3
She can buy 3 bracelets.
Answer:
43560 I think
Step-by-step explanation:
20 x 11 x 11 x 18
Pe eriod b. lud is lea.ving my cgcoo.chie hole.
Answer:

Step-by-step explanation:
The first step is to use the Distance Formula to figure out the lengths of the two legs:
![\sqrt{[-x_1 + x_2]^{2} + [-y_1 + y_2]^{2}} = D](https://tex.z-dn.net/?f=%5Csqrt%7B%5B-x_1%20%2B%20x_2%5D%5E%7B2%7D%20%2B%20%5B-y_1%20%2B%20y_2%5D%5E%7B2%7D%7D%20%3D%20D)
R(1, 1) and P(−1, −2) ↷
![\sqrt{[1 + 1]^{2} + [2 + 1]^{2}} = \sqrt{2^{2} + 3^{2}} = \sqrt{4 + 9} = \sqrt{13}\\ \\ \sqrt{13} = RP](https://tex.z-dn.net/?f=%5Csqrt%7B%5B1%20%2B%201%5D%5E%7B2%7D%20%2B%20%5B2%20%2B%201%5D%5E%7B2%7D%7D%20%3D%20%5Csqrt%7B2%5E%7B2%7D%20%2B%203%5E%7B2%7D%7D%20%3D%20%5Csqrt%7B4%20%2B%209%7D%20%3D%20%5Csqrt%7B13%7D%5C%5C%20%5C%5C%20%5Csqrt%7B13%7D%20%3D%20RP)
Q(2, −4) and P(−1, −2) ↷
![\sqrt{[1 + 2]^{2} + [2 - 4]^{2}} = \sqrt{[-2]^{2} + 3^{2}} = \sqrt{4 + 9} = \sqrt{13} \\ \\ \sqrt{13} = QP](https://tex.z-dn.net/?f=%5Csqrt%7B%5B1%20%2B%202%5D%5E%7B2%7D%20%2B%20%5B2%20-%204%5D%5E%7B2%7D%7D%20%3D%20%5Csqrt%7B%5B-2%5D%5E%7B2%7D%20%2B%203%5E%7B2%7D%7D%20%3D%20%5Csqrt%7B4%20%2B%209%7D%20%3D%20%5Csqrt%7B13%7D%20%5C%5C%20%5C%5C%20%5Csqrt%7B13%7D%20%3D%20QP)
* So, by my calculations, we have an isosceles <em>right triangle</em>, and according to the <em>45°-45°-90° triangle theorem</em>, we automatically know that the hypotenuse, <em>RQ</em>, is
:
30°-60°-90° triangle theorem
2x, x, x√3
↑ ↑ ↑
h leg leg
y
p
o
t
e
n
u
s
e
45°-45°-90° triangle theorem
x√2, x, x
↑ ↑ ↑
h legs
y
p
o
t
e
n
u
s
e
So now, we have to find the area of the triangle by taking half of either height or base, then multiplying that by the height or base, but since this is an isosceles <em>right</em><em> </em><em>triangle</em><em>,</em><em> </em>it does not matter:
![[\frac{h}{2}][b] = A \: \:OR\: \: [\frac{b}{2}][h] = A \: \:OR \: \: \frac{hb}{2} = A](https://tex.z-dn.net/?f=%5B%5Cfrac%7Bh%7D%7B2%7D%5D%5Bb%5D%20%3D%20A%20%5C%3A%20%5C%3AOR%5C%3A%20%5C%3A%20%5B%5Cfrac%7Bb%7D%7B2%7D%5D%5Bh%5D%20%3D%20A%20%5C%3A%20%5C%3AOR%20%5C%3A%20%5C%3A%20%5Cfrac%7Bhb%7D%7B2%7D%20%3D%20A)
![[\frac{\sqrt{13}}{2}][\sqrt{13}] = \frac{13}{2} = 6\frac{1}{2} \\ \\ 6\frac{1}{2} = A](https://tex.z-dn.net/?f=%5B%5Cfrac%7B%5Csqrt%7B13%7D%7D%7B2%7D%5D%5B%5Csqrt%7B13%7D%5D%20%3D%20%5Cfrac%7B13%7D%7B2%7D%20%3D%206%5Cfrac%7B1%7D%7B2%7D%20%5C%5C%20%5C%5C%206%5Cfrac%7B1%7D%7B2%7D%20%3D%20A)
I am joyous to assist you anytime.
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