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3 years ago

I need help i dont understand this,and j have and i have a big test coming up please help thanks

Mathematics

1 answer:

levacccp [35]3 years ago

7 0

Since both of the angles are supplementary, you'd set the expressions up into an equation equal to 180(degrees).
So, your equation should look like:
                       22x + 4 + 35x + 5 = 180.
Next, you combine your like terms so that your next move will look like:
                       57x + 9 = 180.
Then you'd follow the subtraction and division property, so your next moves will look as the following:
                       57x + 9 - 9 = 180 - 9 equals 171
                       57x / 57 = 171 / 57 equals 3
So, finally, your answer is: x = 3.

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3 years ago

Help please tell me what the answer is ASAP

Amiraneli [1.4K]

Answer:

The approximated length of EF is 2.2 units ⇒ A

Step-by-step explanation:

<em>The tangent ratio in the right triangle is the ratio between the opposite side to the adjacent side of one of the acute angle in the triangle</em>

In the given figure

∵ The triangle DFE has a right angle F

∵ The opposite side to angle D is EF

∵ The adjacent side to angle D is DF

→ By using the tangent ratio above

∴ tan(∠D) = $\frac{EF}{DF}$

∵ DF = 6 units

∵ m∠D = 20°

→ Substitute then in the ratio above

∴ tan(20°) = $\frac{EF}{6}$

→ Multiply both sides by 6

∴ 6 tan(20°) = EF

∴ 2.183821406 = EF

→ Approximate it to the nearest tenth

∴ 2.2 = EF

∴ The approximated length of EF is 2.2 units

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3 years ago

How many 10-digit ternary strings are there that contain exactly two 0s, three 1s, and five 2s?

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There are $\dbinom{10}2$ ways of picking 2 of the 10 available positions for a 0. 8 positions remain.

There are $\dbinom83$ ways of picking 3 of the 8 available positions for a 1. 5 positions remain, but we're filling all of them with 2s, and there's $\dbinom55=1$ way of doing that.

So we have

$\dbinom{10}2\dbinom83\dbinom55=\dfrac{10!}{2!(10-2)!}\dfrac{8!}{3!(8-3)!}\dfrac{5!}{5!(5-5)!}=2520$

The last expression has a more compact form in terms of the so-called multinomial coefficient,

$\dbinom{10}{2,3,5}=\dfrac{10!}{2!3!5!}=2520$

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3 years ago