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Gemiola [76]
3 years ago
11

Please help me ASAP thank uu

Mathematics
1 answer:
ZanzabumX [31]3 years ago
4 0

Answer:

51°

Step-by-step explanation:

Remember, all 3 angles must add up to 180°

We can tell that this triangle is a right triangle, or a triangle that has a 90° angle. We also have another angle given, which is 39°. Currently we have two angles given.

Since all 3 angles have to add up to 180°,

90 + 39 + x = 180°

129 + x = 180°

180 - 129 = x

x = 51°

The missing angle is 51°

To see if your calculation is correct, you could add all 3 angles, and it should give you 180 as a sum.

90 + 39 + 51 = 180

Correct!

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Y varies jointly as x, z, and w. when x=2, z=1, and w=12, then y=72. find y when x=1, z=2, and w=3
son4ous [18]
ANSWER: 18


EXPLANATION:

y=kxzw

where k is a constant
Making k the subject of the formula, we have

k=y/xzw

Inputting the values

k= 72/(2)(1)(12)

k=72/24

k = 3

Solving for y when x=1,z=2 and w=3

We have


y=kxzw

y = (3)(1)(2)(3)

y = 18
7 0
2 years ago
1/2(4x+10)=2 i need to get x
kari74 [83]
First you need to distribute the 1/2 by the numbers in the parentheses
1/2•4x and 1/2•10 which equals
2x+5=10

next, you will need to get the 2x by itself so you would do
2x+5=10
-5=-5
2x=5

finally, you divide the 2 by the 5 so you can get the x by itself
2x divided by 5

the answer is: x=2.5
8 0
2 years ago
If y varies directly with x, and y=-1 when x=3
UNO [17]

Answer:

where da question at

Step-by-step explanation:

4 0
3 years ago
(-7)•(-7) • 5•5•5•5 using exponents
kramer

Answer:

what grade level

Step-by-step explanation:

5 0
3 years ago
Please please use explanation with answer
KIM [24]

Answer:

\large\boxed{\theta=-\dfrac{\pi}{4}+k\pi\ or\ \theta=k\pi}\ for\ k\in\mathbb{Z}

Step-by-step explanation:

\tan^2\theta+\tan\theta=0\\\\\tan\theta(\tan\theta+1)=0\iff\tan\theta=0\ \vee\ \tan\theta+1=0\\\\\tan\theta=0\Rightarrow \theta=k\pi\ \text{for}\ k\in\mathbb{Z}\\\\\tan\theta+1=0\qquad\text{subtract 1 from both sides}\\\\\tan\theta=-1\Rightarrow\theta=-\dfrac{\pi}{4}+k\pi\ \text{for}\ k\in\mathbb{Z}

5 0
3 years ago
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