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2 years ago

If m∠D + m∠E = 180°, then ∠D and ∠E are classified as what kind of angles?

1 answer:

3 0

The other person is literally wrong. The answer is D. Supplementary because two angles that add up to 180 are supplementary angles.

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If x=3+root8 and y =3- root 8 find 1/x^2+1/y^2

KonstantinChe [14]

Answer:

Step-by-step explanation:

x=3+√8

y =3-√8

now,

1/x^2+1/y^2

=1/(3+√8)² + 1/(3-√8)²

= [(3-√8)²+(3+√8)²] / (3+√8)²(3-√8)² [L.C.M = (3+√8)²(3-√8)² ]

=[(3-√8+3+√8)²-2(3-√8)(3+√8) ] / [(3+√8)(3-√8)]²

=[6²-2.(3²-√8² )] / (3²-√8²)² [ a²+ b²=(a+b)²-2ab]

=[36-2(9-8) ]/ (9-8)²

=[36-2.1] / 1²

=34

3 0

3 years ago

The expression 21m−49n factored using the GCF

rewona [7]

Here is my work for the problem. Hope this helps!

5 0

3 years ago

Read 2 more answers

O is the midpoint of segment IU. If I is (-7, 2) and O is (5, -8), what are the coordinates of U?

vaieri [72.5K]

Answer:

(17 , -18)

Step-by-step explanation:

5 = (-7 + x) / 2

x = 17

-8 = (2 + y) / 2

y = -18

3 0

2 years ago

toothless car wash (the kid next door) can wash 3 cars in one and a half hours. how long will it take to wash 10 cars

katrin2010 [14]

Answer:

5 hours

Step-by-step explanation:

1.5 (hours) / 3(cars) = 0.5 hours per car

0.5 x 10 = 5

6 0

3 years ago

A candidate for a US Representative seat from Indiana hires a polling firm to gauge her percentage of support among voters in he

mojhsa [17]

Answer:

(A) The minimum sample size required achieve the margin of error of 0.04 is 601.

(B) The minimum sample size required achieve a margin of error of 0.02 is 2401.

Step-by-step explanation:

Let us assume that the percentage of support for the candidate, among voters in her district, is 50%.

(A)

The margin of error, MOE = 0.04.

The formula for margin of error is:

$MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}$

The critical value of z for 95% confidence interval is: $z_{\alpha/2}=1.96$

Compute the minimum sample size required as follows:

$MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}\\0.04=1.96\times \sqrt{\frac{0.50(1-0.50)}{n}}\\(\frac{0.04}{1.96})^{2} =\frac{0.50(1-0.50)}{n}\\n=600.25\approx 601$

Thus, the minimum sample size required achieve the margin of error of 0.04 is 601.

(B)

The margin of error, MOE = 0.02.

The formula for margin of error is:

$MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}$

The critical value of z for 95% confidence interval is: $z_{\alpha/2}=1.96$

Compute the minimum sample size required as follows:

$MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}\\0.02=1.96\times \sqrt{\frac{0.50(1-0.50)}{n}}\\(\frac{0.02}{1.96})^{2} =\frac{0.50(1-0.50)}{n}\\n=2401.00\approx 2401$

Thus, the minimum sample size required achieve a margin of error of 0.02 is 2401.

8 0

3 years ago