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Lena [83]
3 years ago
12

Solve x show your work​

Mathematics
1 answer:
Bezzdna [24]3 years ago
4 0

Answer:

55+x+74+54= 180 degree ( being the sum of all angle of the triangle )

or, 183 +x = 180degree

or, x=(180-183)degree

or, x= -3degree ans

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Staci is attending a music festival. The ticket to the festival costs $87.96 . Staci plans to purchase $30.00 t-shirts from the
mafiozo [28]

Step one: 200-87.96=112.04

Step two: 112.04÷30≈3.7

Step three: 3.7↔3

Answer: Staci can buy 3 shirts

5 0
3 years ago
The solutions to the inequality y > −3x + 2 are shaded on the graph. Which point is a solution?
likoan [24]
"(2, 0)" is the one solution to the inequality y > −3x + 2 among the following choices given in the question that <span>are shaded on the graph. The correct option among all the options that are given in the question is the second option or option "B". I hope that this is the answer that has actually come to your desired help.</span>
8 0
4 years ago
Read 2 more answers
alguien que me ayude con este animaplano esta muy difícil porfa pero sean serios con esto de verdad ​
pashok25 [27]

Answer:

18) 21

19) 18

20) 27

21) 37

22) 38

23) 88

24) 95

25) 87

26)58

27) 77

28) 85

29) 83

30) 64

31) 37

Step-by-step explanation:

Vamos a empezar por sustituyendo las expresiónes

18) 2(^2)+2(3)+3(5)+6(-1)

4+6+15-6

10+11

21

7 0
3 years ago
If cos Θ = square root 2 over 2 and 3 pi over 2 &lt; Θ &lt; 2π, what are the values of sin Θ and tan Θ? sin Θ = square root 2 ov
densk [106]

Answer:

\huge\boxed{\sin\theta=-\dfrac{\sqrt2}{2};\ \tan\theta=-1}

Step-by-step explanation:

We have:

\\cos\theta=\dfrac{\sqrt2}{2},\ \dfrac{3\pi}{2}

For sine use:

\sin^2x+\cos^2x=1\to\sin^2x=1-\cos^2x

Substitute:

\sin^2\theta=1-\left(\dfrac{\sqrt2}{2}\right)^2\\\\\sin^2\theta=1-\dfrac{(\sqrt2)^2}{2^2}\\\\\sin^2\theta=1-\dfrac{2}{4}\\\\\sin^2\theta=\dfrac{4}{4}-\dfrac{2}{4}\\\\\sin^2\theta=\dfrac{4-2}{4}\\\\\sin^2\theta=\dfrac{2}{4}\to\sin\theta=\pm\sqrt{\dfrac{2}{4}}\\\\\sin\theta=\pm\dfrac{\sqrt2}{\sqrt4}\\\\\sin\theta=\pm\dfrac{\sqrt2}{2}

θ in IV quadrant, therefore sine is negative.

\sin\theta=-\dfrac{\sqrt2}{2}

For tangent use:

\tan x=\dfrac{\sin x}{\cos x}

Substitute:

\tan\theta=\dfrac{-\frac{\sqrt2}{2}}{\frac{\sqrt2}{2}}=-\dfrac{\sqrt2}{2}\cdot\dfrac{2}{\sqrt2}=-1

8 0
3 years ago
How to answer h/4 ≤ 2/7
Serjik [45]

I feel you, headaches suck.

Your answer would be  h ≤ 8/7

Step  1  :

           2

Simplify   —

           7

Equation at the end of step  1  :

 h    2

 — -  —  ≤ 0

 4    7

Step  2  :

           h

Simplify   —

           4

Equation at the end of step  2  :

 h    2

 — -  —  ≤ 0

 4    7

Step  3  :

Calculating the Least Common Multiple :

3.1    Find the Least Common Multiple

     The left denominator is :       4

     The right denominator is :       7

After calculating the multipliers you get  h ≤ 8/7

Hope this helps!

Please consider Brainliest!

4 0
3 years ago
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