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dedylja [7]
3 years ago
11

Help please!! Multiple choice geometry coordinates question Question 12

Mathematics
1 answer:
Ghella [55]3 years ago
3 0

In the attached picture you find the two triangles. To check if the lengths of the sides are the same, we use the formula for the distance between two points:

d(A,B) = \sqrt{(A_x-B_x)^2+(A_y-B_y)^2}

If we plug the values for the points of PQR, we have

\overline{PQ} \approx 5.39

\overline{QR} \approx 3.61

\overline{PR} \approx 7.07

Similarly, for STU we have

\overline{ST} \approx 5.1

\overline{TU} \approx 2.24

\overline{SU} \approx 6.71

As you can see, the lengths of the sides do not match, as a triplet, so the two triangles are not congruent.

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If a=2b^3 and b= -1/2c ^-2,express a in terms of c.
ElenaW [278]

Answer:

Part 1) a=-\frac{1}{4c^6}

Part 2) a=-\frac{1}{4c^{-6}}

Step-by-step explanation:

Part 1) we have

a=2b^{3} ----> equation A

b=-\frac{1}{2}c^{-2} ----> equation B

substitute equation B in equation A

a=2(-\frac{1}{2}c^{-2})^{3}

Applying property of exponents

(x^{m})^{n}=x^{m*n}

x^{-m} =\frac{1}{x^{m}}

a=2(-\frac{1}{2}c^{-2})^{3}=2(-\frac{1}{2})^3(c^{-2})^{3}=2(-\frac{1}{8})(c^{-6})=-\frac{1}{4c^6}

therefore

a=-\frac{1}{4c^6}

Part 2) we have

a=2b^{3} ----> equation A

b=-\frac{1}{2c^{-2}}=-\frac{c^{2}}{2} ----> equation B

substitute equation B in equation A

a=2(-\frac{c^{2}}{2})^{3}

Applying property of exponents

(x^{m})^{n}=x^{m*n}

x^{-m} =\frac{1}{x^{m}}

a=2(-\frac{c^{2}}{2})^{3}=2(-\frac{c^{6}}{8})

simplify

a=-\frac{c^{6}}{4}

therefore

a=-\frac{1}{4c^{-6}}

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Answer:

ST=10

Step-by-step explanation:

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C. This is a claim

D. This is rejection of a claim that mean is 1 pound

E. This is rejection of claim that average delivery time is 3 days.

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