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Natasha2012 [34]
4 years ago
7

Can someone please help me with this problem :)

Mathematics
1 answer:
polet [3.4K]4 years ago
3 0

Answer:

Tan A = x/8

The portion facing the right angle is always the opposite in any triangle.

Tan is gotten from TOA

TAN=OPP/ADJ

OPP=X

ADJ=8

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How to solve 30+(4x+2)=8+6x)
Gennadij [26K]

Answer:

12

Step-by-step explanation:

30+(4x+2)=8+6x

30+4x+2=8+6x

32+4x=8+6x

32=8+6x-4x

32=8+2x

2x=32-8

2x=24

x=24/2

x=12

4 0
4 years ago
Read 2 more answers
Select the descriptions that are true of the equation y - 6= - 3(x+2). Select all that apply.
Lapatulllka [165]

The point-slope form:

y-y_1=m(x-x_1)

We have

y-6=-3(x+2)\to y-6=-3(x-(-2))

Therefore

<h3>slope = -3 and the point (-2, 6)</h3>

------------

<h3>y + 2 = 2(x - 1)</h3>

2x - y = 4    <em>it's standard form Ax + By = C</em>

y = 2x - 4     <em>it's slope-intercept form y = mx + b</em>

y - x = 4 →<em> </em>x - y = -4  <em>t's standard form Ax + By = C</em>

3 0
4 years ago
I need help, ive been trying
Mrac [35]

The figure is a right triangle so we can use the Pythagorean theorem to find the third side AKA the hypotenuse.

Pythagorean theorem = a² + b² = c² where A and B are the legs of the triangle and C is the hypotenuse.

39² + 80² = c²

169 + 6400 = c²

6569 = c²

Square root both sides to cancel out the squared variable.

√6569 = √(c²)

√6569 = c

The third side, the hypotenuse, is √6569 units long.

7 0
3 years ago
NEEED HELP FROM BRAINLIEST??????
Leno4ka [110]

Answer:

A

Step-by-step explanation:

5 0
3 years ago
In a box there are six envelopes each containing two cards. Three of the envelopes contain two red cards, two of them contain a
Sphinxa [80]

Answer:

\frac{2}{5}

Step-by-step explanation:

3 envelopes having 2 red card

2 envelopes having 1 red card and 1 black card

1 envelope having 2 black cards

We are given that . An envelope is selected at random and a card is withdrawn and found to be red.

So, No. of ways of envelope having red card = 3+2 = 5

No. of required ways of envelope having 1 red card and 1 black card = 2

So, probability of getting an envelope having 1 red card and 1 black card = \frac{2}{5}

Hence The chance the other card is black is \frac{2}{5}

5 0
3 years ago
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