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alekssr [168]
3 years ago
5

Caitlyn did 6/7 of the math quiz

Mathematics
1 answer:
lord [1]3 years ago
6 0
What is this supposed to mean?
You might be interested in
-6 + 3x = 9 <br> Solve for x
Eduardwww [97]

Answer:

5

Step-by-step explanation:

3x = 15 (you add the 6 over)

x = 5 (you divide by 3)

4 0
3 years ago
Solve the missing side. Find the missing side. Round to the nearest tenth. Please show the work. Part 2. #5-8​
enyata [817]

Answer:

5. 6.1 units

6. 29.4 units

7. 12.8 units

8. 29.1 units

Step-by-step explanation:

5. We have the opposite and adjacent sides here for angle 21 degrees. So we need to use tangent, which is opposite / adjacent:

tan(21)=x/16

x=16*tan(21) ≈ 6.1 units

6. We have the opposite and hypotenuse sides here for angle 33 degrees. So we need to use sine, which is opposite / hypotenuse:

sin(33)=16/x

x=16/sin(33) ≈ 29.4 units

7. We have the opposite and adjacent sides here for angle 37 degrees. So we need to use tangent, which is opposite / adjacent:

tan(37)=x/17

x=17*tan(37) ≈ 12.8 units

8. We have the opposite and hypotenuse sides here for angle 31 degrees. So we need to use sine, which is opposite / hypotenuse:

sin(31)=15/x

x=15/sin(31) ≈ 29.1 units

Hope this helps!

4 0
3 years ago
Read 2 more answers
In the figure, a∥e , m∥n , and m∠3 = 108°. What is the m∠6 ? Enter your answer in the box.
maks197457 [2]
I believe that angle 6 is 108 because I drew to the best of my ability and I see that 3 and 6 are alternate interior angles.
3 0
3 years ago
This is pretty hard but i really need it correct please help!
Wewaii [24]
I prob think it’s b
6 0
3 years ago
Please help me with this problem
aev [14]

Answer:  \bold{b)\quad \dfrac{17\sqrt2}{26}}

<u>Step-by-step explanation:</u>

\text{Given }sin\ x=-\dfrac{5}{13}\quad \text{and is in Quadrant IV}\quad \implies\quad cos\ x=\dfrac{12}{13}\\\\\text{Use the cosine sum formula: }cos(A + B) = cosA\cdot cosB-sinA\cdot sinB\\\\\implies cos\bigg(x+\dfrac{\pi}{4}\bigg)=cosx\cdot cos\dfrac{\pi}{4}-sinx\cdot sin\dfrac{\pi}{4}\\\\\\=\dfrac{12}{13}\cdot\dfrac{\sqrt2}{2}-\bigg(-\dfrac{5}{13}\bigg)\cdot \dfrac{\sqrt2}{2}\\\\\\=\dfrac{12\sqrt2}{26}+\dfrac{5\sqrt2}{26}\\\\\\=\large\boxed{\dfrac{17\sqrt2}{26}}

8 0
4 years ago
Read 2 more answers
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