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Tomtit [17]
3 years ago
15

1/8 divided by 3/4 please help

Mathematics
2 answers:
zimovet [89]3 years ago
8 0
1/6 Hoped I helped ;)
dlinn [17]3 years ago
6 0
1/8 divided by 3/4 is (1/8) / (3/4). When you divide fractions, you flip the 2nd number and change the divide into multiply. So (1/8)*(4/3). Multiply top by top and bottom by bottom to get 4/24. That simplifies to 1/6
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Please help I’m struggling
gulaghasi [49]

Answer: B, x=20; angle measure is 30°.

Step-by-step explanation:

Opposite angles are always the same.

You can see that 30° and 3x are opposite angles therefore 3x=30°.

In algebra, when a letter and a number are next to each other, it means times.

So, 3x=30° means 3 times something equals 30.

And we know that 3×10=30 so, x=10.

Hope this helps :)

3 0
3 years ago
I need some help with this, please explain well. It's Trigonometry with right triangles, please let me know if you need a better
andreyandreev [35.5K]

Answer:

x = 12

Step-by-step explanation:

First you figure out which function your going to use. I notice that from the angle we have 5 which is the opposite side and 23 which is the adjacent side. This means we have to use tangent.

Then set up your equation: tan(x)=5/23. This is because the opposite side is 5 and the adjacent is 23 and it must be set up in that order.

Next get x alone by making the equation: x=tan^-1(5/23).

Then put it into the calculator and round to get 12.

5 0
4 years ago
Factor the expression<br> 7x^2 + 50x + 7<br> (Type your answer in factored form.)
alina1380 [7]

Answer: (7x + 1)(x + 7)

Step-by-step explanation:

$$Factor the following:$$7 x^{2}+50 x+7$$Factor the quadratic $7 x^{2}+50 x+7$. The coefficient of $x^{2}$ is 7 and the constant term is 7. The product of 7 and 7 is 49 . The factors of 49 which sum to 50 are 1 and 49. So:\\\\ $7 x^{2}+50 x+7\\=7 x^{2}+49 x+x+7\\=7(7 x+1)+x(7 x+1)$ \\\\Factor $7 x+1$ from $7(7 x+1)+x(7 x+1)$ :$$\\\\Answer:(7 x+1)(x+7)$$

3 0
2 years ago
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8 0
3 years ago
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Mrrafil [7]

Answer:

true

Step-by-step explanation:

5 0
3 years ago
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