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Veronika [31]
4 years ago
12

Line M is parallel to line L. Name the angles which are congruent to angle C in the figure shown below:

Mathematics
2 answers:
insens350 [35]4 years ago
8 0

A) S, Q, and B

Hope that help you!

tino4ka555 [31]4 years ago
8 0

Answer:

S, Q and B

Step-by-step explanation:

Those are the angles.

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Lowering powers write in terms of first power of cosine. Cos^6
yaroslaw [1]

The main identity you need is the double angle one for cosine:

\cos^2x=\dfrac{1+\cos2x}2

We get

\cos^6x=(\cos^2x)^3=\left(\dfrac{1+\cos2x}2\right)^3=\dfrac{(1+\cos2x)^3}8

Expand the numerator to apply the identity again:

\cos^6x=\dfrac{1+3\cos2x+3\cos^22x+\cos^32x}8

\cos^6x=\dfrac{1+3\cos2x+3\left(\frac{1+\cos2(2x)}2\right)+\cos2x\left(\frac{1+\cos2(2x)}2\right)}8

\cos^6x=\dfrac{1+3\cos2x+\frac32+\frac32\cos4x+\frac12\cos2x(1+\cos4x)}8

\cos^6x=\dfrac5{16}+\dfrac7{16}\cos2x+\dfrac3{16}\cos4x+\dfrac1{16}\cos2x\cos4x

Finally, make use of the product identity for cosine:

\cos2x\cos4x=\dfrac{\cos6x+\cos2x}2

so that ultimately,

\cos^6x=\dfrac5{16}+\dfrac7{16}\cos2x+\dfrac3{16}\cos4x+\dfrac1{32}\cos2x+\dfrac1{32}\cos6x

\cos^6x=\dfrac5{16}+\dfrac{15}{32}\cos2x+\dfrac3{16}\cos4x+\dfrac1{32}\cos6x

4 0
4 years ago
PLEAS HELP QUICK What is the volume of the figure?
mrs_skeptik [129]

Answer:

1530 cubic cm

Step-by-step explanation:

(18*4*10) + (9*9*10) = 1530

4 0
3 years ago
If (a, –5) is a solution to the equation 3a = –2b – 7, what is a?
sukhopar [10]

Answer:

d)  1

Step-by-step explanation:

3a = –2b – 7

We have the point (a, -5) so b = -5

Substituting in

3a = -2(-5) -7

3a = 10-7

3a = 3

Divide each side by 3

3a/3 = 3/3

a =1

6 0
3 years ago
based only on the information given in the diagram which congruence theorems or postulates could be given as reason why ABC=DEF
oksian1 [2.3K]

Answer:

HL

SAS

SSS

Step-by-step explanation:

Since these are right triangles, and you have the two hypotenuses are congruent to each other and two legs that are also congruent to each other, then HL can be applied.

For HA to work we must have been given something else about one of those angle (besides the 90 degree one).

Since you have two corresponding sides that are congruent, then the 90 degree angles in both are congruent, and then the sides right after that 90 degree angle are also congruent to each other, so SAS can be applied.

We can't use AAS.  We only know something about one angle per each triangle due to the markers.

LA? Needed another angle besides the 90 degree one.

All three corresponding sides are congruent. The markers tell us this. So we can apply SSS.

7 0
4 years ago
Solve this polynomial:
ivann1987 [24]
2d^2 -2d-7

(14d^2 - 8) + (6d^2 - 2d + 1)
14d^2 - 8+ 6d^2 - 2d + 1
20d^2 - 7 - 2d
20d^2 - 2d - 7
4 0
3 years ago
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