Solve the right triangle, AABC, for the missing sides and angle to

2. In a school with only sixth and seventh graders

inna [77]

Answer:

there will be 300 6th graders

Step-by-step explanation:

Divide 200 by 2 to find 1/5 of the school.

5 0

3 years ago

Please help with the attachment

just olya [345]

Answer:

b

Step-by-step explanation:

In y=mx+b form, the b is the y-intercept, therefore where the line crosses the y-axis on the graph. In this graph, the line crosses the y-axis at 100. This means that the "b" value is 100. So far, our equation is y=mx+100.

Next, to find m, you go to any other point on the graph and see how much units (rise over run) you went to get there. I picked the point (2,150), You can see that it took me 1 unit up and 2 units to the right to get to that point, leading us to believe the slope is 1/2.

Finally, put everything together and you get the answer of y=1/2x+100 (the letters do not matter) :)

6 0

3 years ago

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Simplify 4 + (-3 -8)

const2013 [10]

Answer:

-7

Step-by-step explanation:

4 0

4 years ago

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1. S(–4, –4), P(4, –2), A(6, 6) and Z(–2, 4) a) Apply the distance formula for each side to determine whether SPAZ is equilatera

Aleksandr [31]

Answer:

a) SPAZ is equilateral.

b) Diagonals SA and PZ are perpendicular to each other.

c) Diagonals SA and PZ bisect each other.

Step-by-step explanation:

At first we form the triangle with the help of a graphing tool and whose result is attached below. It seems to be a paralellogram.

a) If figure is equilateral, then SP = PA = AZ = ZS:

$SP = \sqrt{[4-(-4)]^{2}+[(-2)-(-4)]^{2}}$

$SP \approx 8.246$

$PA = \sqrt{(6-4)^{2}+[6-(-2)]^{2}}$

$PA \approx 8.246$

$AZ =\sqrt{(-2-6)^{2}+(4-6)^{2}}$

$AZ \approx 8.246$

$ZS = \sqrt{[-4-(-2)]^{2}+(-4-4)^{2}}$

$ZS \approx 8.246$

Therefore, SPAZ is equilateral.

b) We use the slope formula to determine the inclination of diagonals SA and PZ:

$m_{SA} = \frac{6-(-4)}{6-(-4)}$

$m_{SA} = 1$

$m_{PZ} = \frac{4-(-2)}{-2-4}$

$m_{PZ} = -1$

Since $m_{SA}\cdot m_{PZ} = -1$ , diagonals SA and PZ are perpendicular to each other.

c) The diagonals bisect each other if and only if both have the same midpoint. Now we proceed to determine the midpoints of each diagonal:

$M_{SA} = \frac{1}{2}\cdot S(x,y) + \frac{1}{2}\cdot A(x,y)$

$M_{SA} = \frac{1}{2}\cdot (-4,-4)+\frac{1}{2}\cdot (6,6)$

$M_{SA} = (-2,-2)+(3,3)$

$M_{SA} = (1,1)$

$M_{PZ} = \frac{1}{2}\cdot P(x,y) + \frac{1}{2}\cdot Z(x,y)$

$M_{PZ} = \frac{1}{2}\cdot (4,-2)+\frac{1}{2}\cdot (-2,4)$

$M_{PZ} = (2,-1)+(-1,2)$

$M_{PZ} = (1,1)$

Then, the diagonals SA and PZ bisect each other.

8 0

3 years ago

A high school sewing club has 11 members. In how many different ways can they choose a president and vice president?

lorasvet [3.4K]

Since the order matters Person A and B can be chosen as AB (A pres B vice pres) or BA (B president A Vice president)

When the order matters we are talking about permutations. How many permutations of 11 taken 2 at a time are there.
It can be written like this 11nPr 2
It means start at 11 and multiply by the next lower number
11 x 10 = 110 ways they can choose.
Extension of the concept:
If there had been three positions President, Vice President, and Treasurer
then it would be 11nPr3 or 11 x 10 x 9
Once again the answer to your problem is 110.

6 0

4 years ago

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