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3 years ago

Solve for X Explain if possible :)

Mathematics

1 answer:

sweet-ann [11.9K]3 years ago

4 0

The two labeled angles are alternate interior angles, and as such, they are the same.

From this result you can build the equation

$15x-2 = 13x+2$

and solve it for x: subtract 13x from both sides to get

$2x-2=2$

and add 2 to both sides to get

$2x = 4 \implies x = 2$

Check: if we plug the value we found we have

$13\cdot 2 + 2 = 26+2 = 28 = 15\cdot 2 - 2 = 30-2$

So the angles are actually the same, as requested.

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The quotient of 2 and a number x, times 3

alexira [117]

Answer:

3(x + 2)

Step-by-step explanation:

$quotient = (x + 2) \times 3 \\ \hspace{42 pt} = 3(x + 2)$

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3 years ago

I WILL MARK BRAINLIEST

never [62]

Answer:

Step-by-step explanation:

The tuna can has a shape of a cylinder, and the area of the tuna lablel is the lateral surface area of the can

$A=2\pi rh$

Data

diameter d= 6.5cm

radius r = $\frac{diameter}{2}$ = $\frac{6.5}{2} = 3.25cm$

height h= 3cm

leteral surface area

$A=2*3.142*3.25*3\\= 61.269\\=61.27cm^2$

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3 years ago

A soccer field is 100 meters long. What could the length in yards

Zolol [24]

Answer:

109 yards.

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

A girl obtained 32.8 marks in nepali and failed by 7.2 marks. What was the pass marks of the subject?

adoni [48]

Answer:

40 marks was the pass marks of the subject.

Step-by-step explanation:

32.8 + 7.2 = 40 marks { basic addition theory based }

8 0

2 years ago

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