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2 years ago

Which expression is equivalent to the expression below?

Mathematics

2 answers:

yaroslaw [1]2 years ago

5 0

Answer:

Step-by-step explanation:

I think it is this answer

Elenna [48]2 years ago

5 0

Answer:

g+g+g+g+g+g=6g

Step-by-step explanation:

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Solve for x. Round to the nearest tenth of a degreee, if necessary

damaskus [11]

Answer:

Step-by-step explanation:

Tan 0= opposite÷ adjacent

Tan 0= 6.7 ÷ 3.4

Tan 0 = 1.97

V= 63.09

V = 63

8 0

2 years ago

Help!!! Algebra1B

Andreyy89

Answer:

3) y = 2

Step-by-step explanation:

Step : 1

given 2 y-1 =3

Adding '1 ' on both sides, we get

2 y -1 +1 = 3+1

2 y = 4

dividing by '2' on both sides, we get

$\frac{2y}{2} = \frac{4}{2}$

cancelling '2' on both sides, we get

y = 2

7 0

3 years ago

What's 2.537 rounded to the nearest hundred?

Mnenie [13.5K]

Answer: 2.54

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

If 180° < α < 270°, cos⁡ α = −817, 270° < β < 360°, and sin⁡ β = −45, what is cos⁡ (α + β)?

eduard

Answer:

$cos(\alpha+\beta)=-\frac{84}{85}$

Step-by-step explanation:

we know that

$cos(\alpha+\beta)=cos(\alpha)*cos(\beta)-sin(\alpha)*sin(\beta)$

Remember the identity

$cos^{2} (x)+sin^2(x)=1$

step 1

Find the value of $sin(\alpha)$

we have that

The angle alpha lie on the III Quadrant

The values of sine and cosine are negative

$cos(\alpha)=-\frac{8}{17}$

Find the value of sine

$cos^{2} (\alpha)+sin^2(\alpha)=1$

substitute

$(-\frac{8}{17})^{2}+sin^2(\alpha)=1$

$sin^2(\alpha)=1-\frac{64}{289}$

$sin^2(\alpha)=\frac{225}{289}$

$sin(\alpha)=-\frac{15}{17}$

step 2

Find the value of $cos(\beta)$

we have that

The angle beta lie on the IV Quadrant

The value of the cosine is positive and the value of the sine is negative

$sin(\beta)=-\frac{4}{5}$

Find the value of cosine

$cos^{2} (\beta)+sin^2(\beta)=1$

substitute

$(-\frac{4}{5})^{2}+cos^2(\beta)=1$

$cos^2(\beta)=1-\frac{16}{25}$

$cos^2(\beta)=\frac{9}{25}$

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

step 3

Find cos⁡ (α + β)

$cos(\alpha+\beta)=cos(\alpha)*cos(\beta)-sin(\alpha)*sin(\beta)$

we have

$cos(\alpha)=-\frac{8}{17}$

$sin(\alpha)=-\frac{15}{17}$

$sin(\beta)=-\frac{4}{5}$

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

substitute

$cos(\alpha+\beta)=-\frac{8}{17}*\frac{3}{5}-(-\frac{15}{17})*(-\frac{4}{5})$

$cos(\alpha+\beta)=-\frac{24}{85}-\frac{60}{85}$

$cos(\alpha+\beta)=-\frac{84}{85}$

4 0

3 years ago

NEED HELP IN MATH , SERIOUS ANSWERS PLEASE

Damm [24]

The answer would be they are congruent.

It's because there was no vertical/horizontal stretch and compression listed in the problem's transformations. The figure was translated throughout the graph.

4 0

2 years ago

Read 2 more answers