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2 years ago

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Which quadrilateral does NOT always have two pairs of parallel sides? A. A Square B. A Parallelogram C. A Trapezoid D. A Rhombus

5th grade math

2 answers:

Yuliya22 [10]2 years ago

4 0

D a rhombus because it dosnt always have congruent sides

lilavasa [31]2 years ago

4 0

Answer:d a rhombus because it dosnt always have congruent sides

Step-by-step explanation:

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What is the measure of angel A rounded to nearest whole number ?

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If A and B are two angles in standard position in Quadrant I, find cos( A +B ) for the given function values. sin A = 8/17 and c

horsena [70]

Answer:

Part 1) $cos(A + B) = \frac{140}{221}$

Part 2) $cos(A - B) = \frac{153}{185}$

Part 3) $cos(A - B) = \frac{84}{85}$

Part 4) $cos(A + B) = -\frac{36}{85}$

Part 5) $cos(A - B) = \frac{63}{65}$

Part 6) $cos(A+ B) = -\frac{57}{185}$

Step-by-step explanation:

<u><em>the complete answer in the attached document</em></u>

Part 1) we have

$sin(A)=\frac{8}{17}$

$cos(B)=\frac{12}{13}$

Determine cos (A+B)

we know that

$cos(A + B) = cos(A) cos(B)-sin(A) sin(B)$

step 1

Find the value of cos(A)

Remember that

$cos^2(A)+sin^2(A)=1$

substitute the given value

$cos^2(A)+(\frac{8}{17})^2=1$

$cos^2(A)+\frac{64}{289}=1$

$cos^2(A)=1-\frac{64}{289}$

$cos^2(A)=\frac{225}{289}$

$cos(A)=\pm\frac{15}{17}$

The angle A belong to the I quadrant, the cosine is positive

$cos(A)=\frac{15}{17}$

step 2

Find the value of sin(B)

Remember that

$cos^2(B)+sin^2(B)=1$

substitute the given value

$sin^2(B)+(\frac{12}{13})^2=1$

$sin^2(B)+\frac{144}{169}=1$

$sin^2(B)=1-\frac{144}{169}$

$sin^2(B)=\frac{25}{169}$

$sin(B)=\pm\frac{25}{169}$

The angle B belong to the I quadrant, the sine is positive

$sin(B)=\frac{5}{13}$

step 3

Find cos(A+B)

substitute in the formula

$cos(A + B) = \frac{15}{17} \frac{12}{13}-\frac{8}{17}\frac{5}{13}$

$cos(A + B) = \frac{180}{221}-\frac{40}{221}$

$cos(A + B) = \frac{140}{221}$

Part 2) we have

$sin(A)=\frac{3}{5}$

$cos(B)=\frac{12}{37}$

Determine cos (A-B)

we know that

$cos(A - B) = cos(A) cos(B)+sin(A) sin(B)$

step 1

Find the value of cos(A)

Remember that

$cos^2(A)+sin^2(A)=1$

substitute the given value

$cos^2(A)+(\frac{3}{5})^2=1$

$cos^2(A)+\frac{9}{25}=1$

$cos^2(A)=1-\frac{9}{25}$

$cos^2(A)=\frac{16}{25}$

$cos(A)=\pm\frac{4}{5}$

The angle A belong to the I quadrant, the cosine is positive

$cos(A)=\frac{4}{5}$

step 2

Find the value of sin(B)

Remember that

$cos^2(B)+sin^2(B)=1$

substitute the given value

$sin^2(B)+(\frac{12}{37})^2=1$

$sin^2(B)+\frac{144}{1,369}=1$

$sin^2(B)=1-\frac{144}{1,369}$

$sin^2(B)=\frac{1,225}{1,369}$

$sin(B)=\pm\frac{35}{37}$

The angle B belong to the I quadrant, the sine is positive

$sin(B)=\frac{35}{37}$

step 3

Find cos(A-B)

substitute in the formula

$cos(A - B) = \frac{4}{5} \frac{12}{37}+\frac{3}{5} \frac{35}{37}$

$cos(A - B) = \frac{48}{185}+\frac{105}{185}$

$cos(A - B) = \frac{153}{185}$

Part 3) we have

$sin(A)=\frac{15}{17}$

$cos(B)=\frac{3}{5}$

Determine cos (A-B)

we know that

$cos(A - B) = cos(A) cos(B)+sin(A) sin(B)$

step 1

Find the value of cos(A)

Remember that

$cos^2(A)+sin^2(A)=1$

substitute the given value

$cos^2(A)+(\frac{15}{17})^2=1$

$cos^2(A)+\frac{225}{289}=1$

$cos^2(A)=1-\frac{225}{289}$

$cos^2(A)=\frac{64}{289}$

$cos(A)=\pm\frac{8}{17}$

The angle A belong to the I quadrant, the cosine is positive

$cos(A)=\frac{8}{17}$

step 2

Find the value of sin(B)

Remember that

$cos^2(B)+sin^2(B)=1$

substitute the given value

$sin^2(B)+(\frac{3}{5})^2=1$

$sin^2(B)+\frac{9}{25}=1$

$sin^2(B)=1-\frac{9}{25}$

$sin^2(B)=\frac{16}{25}$

$sin(B)=\pm\frac{4}{5}$

The angle B belong to the I quadrant, the sine is positive

$sin(B)=\frac{4}{5}$

step 3

Find cos(A-B)

substitute in the formula

$cos(A - B) = \frac{8}{17} \frac{3}{5}+\frac{15}{17} \frac{4}{5}$

$cos(A - B) = \frac{24}{85}+\frac{60}{85}$

$cos(A - B) = \frac{84}{85}$

Part 4) we have

$sin(A)=\frac{15}{17}$

$cos(B)=\frac{3}{5}$

Determine cos (A+B)

we know that

$cos(A + B) = cos(A) cos(B)-sin(A) sin(B)$

step 1

Find the value of cos(A)

Remember that

$cos^2(A)+sin^2(A)=1$

substitute the given value

$cos^2(A)+(\frac{15}{17})^2=1$

$cos^2(A)+\frac{225}{289}=1$

$cos^2(A)=1-\frac{225}{289}$

$cos^2(A)=\frac{64}{289}$

$cos(A)=\pm\frac{8}{17}$

The angle A belong to the I quadrant, the cosine is positive

$cos(A)=\frac{8}{17}$

step 2

Find the value of sin(B)

Remember that

$cos^2(B)+sin^2(B)=1$

substitute the given value

$sin^2(B)+(\frac{3}{5})^2=1$

$sin^2(B)+\frac{9}{25}=1$

$sin^2(B)=1-\frac{9}{25}$

$sin^2(B)=\frac{16}{25}$

$sin(B)=\pm\frac{4}{5}$

The angle B belong to the I quadrant, the sine is positive

$sin(B)=\frac{4}{5}$

step 3

Find cos(A+B)

substitute in the formula

$cos(A + B) = \frac{8}{17} \frac{3}{5}-\frac{15}{17} \frac{4}{5}$

$cos(A + B) = \frac{24}{85}-\frac{60}{85}$

$cos(A + B) = -\frac{36}{85}$

4 0

3 years ago

Which statements are true about reflections? Check all that apply.

maria [59]

Answer:

A, B, C, D, F

Step-by-step explanation:

<em>A. An image created by a reflection will always be congruent to its pre-image. </em>

Congruent means identical or similar. The orientation, size, and shape of the pre-image are not changed after a reflection- therefore, the pre-image and image are congruent.

<em>B. An image and its pre-image are always the same distance from the line of reflection.</em>

When a pre-image is flipped across the line of reflection, both figures are perfectly symmetrical. Nothing has changed aside from orientation and location on the graph. The number of units between the figures and the line of reflection should remain unchanged.

<em>C. If a point on the pre-image lies on the line of reflection, the image of that point is the same as the pre-image.</em>

When you reflect a pre-image with a point that lies on the line of reflection, that point will stay on the line of the reflection as the rest of the figure is flipped.

I will explain these two answers together since they have similar explanations.

<em>D. The line of reflection is perpendicular to the line segments connecting corresponding vertices.</em>

<em>F. The line segments connecting corresponding vertices are all parallel to each other.</em>

In a reflection, the horizontal or vertical (depending on the line of reflection, y-axis or x-axis) coordinates are altered. However, if one were to draw an imaginary line connecting the corresponding vertices, both the image and pre-image are parallel and perpendicular.

I hope this helped, good luck on your work/test!

7 0

2 years ago