All categories

3 years ago

What is the value of x ?

Mathematics

1 answer:

Lostsunrise [7]2 years ago

3 0

Vertical angles are congruent (equal), and the angles below are vertical angles.

x + 40 = 3x

40 = 3x - x

2x = 40

You might be interested in

Which inequality would be in the correct slope-intercept form:

WINSTONCH [101]

Answer:

J. y > 3x + 5

Step-by-step explanation:

To convert the standard form equation into slope-intercept, we must make y isolated.

Step 1: Subtract 6x from both sides.

$6x-2y-6x$
$-2y$

Step 2: Divide both sides by -2 and flip the inequality sign.

$-2y/-2 < (-6x-10)/-2$
$y>3x + 5$

Therefore, the answer is J.

Have a lovely rest of your day/night, and good luck with your assignments! ♡

3 0

2 years ago

Read 2 more answers

3) Round answer to the nearest hundredth. *

DochEvi [55]

Answer:

80 mm

Step-by-step explanation:

Pythagorean theorm can be applied here to solve for the length of the last leg, $a^{2}$ + $b^{2}$ = $c^{2}$ , where (a) and (b) are the legs of the triangle and (c) is the hypotenuse. We can rearrange the pythagorean theorm for our purposes like the following:

b = √c^2 - √a^2

= √(82)^2 - √(18)^2

= √6724 - √324

= √6400

= 80 mm

5 0

2 years ago

Meghan and Sabrina compared the amount of interest they each earned on their savings accounts. Each had deposited $1000, but Meg

Margarita [4]

Sabrina because she earned more money then meghen

3 0

2 years ago

Read 2 more answers

PLEASE HELP! BRAINLIEST!

balu736 [363]

Answer:

x=4

Step-by-step explanation:

We can find the length of RT by using the Pythagorean theorem

sin theta = opposite side/ hypotenuse

sin 60 = 2 sqrt(3)/ RT

Multiply each side by RT

RT sin 60 = 2 sqrt(3)

Divide by sin 60

RT = 2 sqrt(3)/ sin 60

RT = 4

Then

tan theta = opposite side/ adjacent side

tan 45 = x/RT

tan 45 = x/4

4 tan 45 = x

4 = x

6 0

2 years ago

Tan^2 A/1+cot^2 A + cot^2 A/1+tan^2 A=sec^2 A cosec^2 A-3

omeli [17]

$\frac{tan^2x}{1+cot^2x}+\frac{cot^2x}{1+tan^2x}=sec^2x\ cosec^2x-3\\\\L=\frac{tan^2x(1+tan^2x)+cot^2x(1+cot^2x)}{(1+cot^2x)(1+tan^2x)}=\frac{tan^2x+tan^4x+cot^2x+cot^4x}{1+tan^2x+cot^2x+tanxcotx}\\\\=\frac{tan^2x+cot^2x+tan^4x+cot^4x}{1+tan^2x+cot^2x+1}=\frac{tan^2x+2+cot^2x+tan^4x-2+cot^4x}{tan^2x+cot^2x+2}$

$=\frac{(tanx+cotx)^2+(tan^2x-cot^2x)^2}{(tanx+cotx)^2}=\frac{(tanx+cotx)^2}{(tanx+cotx)^2}+\frac{(tan^2x-cot^2x)^2}{(tanx+cotx)^2}\\\\=1+\frac{(tanx-cotx)^2(tanx+cotx)^2}{(tanx+cotx)^2}=1+(tanx-cotx)^2\\\\=1+tan^2x-2tanx\ cotx+cot^2x=tan^2x+cot^2x+1-2\\\\=\left(\frac{sinx}{cosx}\right)^2+\left(\frac{cosx}{sinx}\right)^2-1=\frac{sin^2x}{cos^2x}+\frac{cos^2x}{sin^2x}-1=\frac{sin^4x+cos^4x}{sin^2x\ cos^2x}-1$

$=\frac{(sin^2x)^2+2sin^2x\ cos^2x+(cos^2x)^2-2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1\\\\=\frac{(sin^2x+cos^2x)^2-2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1=\frac{1^2-2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1\\\\=\frac{1}{sin^2x\ cos^2x}-\frac{2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1=\frac{1}{sin^2x}\cdot\frac{1}{cos^2x}-2-1\\\\=cosec^2x\cdot sec^2x-3=sec^2x\ cosec^2x-3=R$

3 0

3 years ago