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2 years ago

Please help I’m struggling

Mathematics

1 answer:

gulaghasi [49]2 years ago

3 0

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 :)

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Please help me with this question with full solutions!!!

Novay_Z [31]

Answer: Choice C

x/w and z/(y+v)

======================================================

Explanation:

All answer choices have that first fraction with a denominator of w. This implies that AB = w is the hypotenuse. This only applies to triangle ABD.

Focus on triangle ABD. It has an opposite leg of AD = x, when the reference angle is ABD (or angle B for short).

So we can say sin(ABD) = opposite/hypotenuse = AD/AB = x/w

x/w is one of the answers

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Also note how y+v is the same for each denominator in the second fraction. y+v is the hypotenuse of triangle ABC. When the reference angle is ABD (aka angle ABC), the opposite side of this same triangle is AX = z

Therefore,

sin(ABD) = sin(ABC) = opp/hyp = AC/BC = z/(y+v)

z/(y+v) is the other answer

Side note: triangle ACD is not used at all.

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3 years ago

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What is a GCF of 10 and 20

USPshnik [31]

The greatest common factor of 10 and 20 is 10.

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3 years ago

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What is 2x +3y+6x+9y

shusha [124]

Combine like terms to get 2x + 6x which is 8x and 3y + 9y which is 12y so the simplified answer is 8x + 12y

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3 years ago

Prove the following DeMorgan's laws: if LaTeX: XX, LaTeX: AA and LaTeX: BB are sets and LaTeX: \{A_i: i\in I\} {Ai:i∈I} is a fam

MariettaO [177]

$X-(A\cup B)=(X-A)\cap(X-B)$

I'll assume the usual definition of set difference, $X-A=\{x\in X,x\not\in A\}$ .

Let $x\in X-(A\cup B)$ . Then $x\in X$ and $x\not\in(A\cup B)$ . If $x\not\in(A\cup B)$ , then $x\not\in A$ and $x\not\in B$ . This means $x\in X,x\not\in A$ and $x\in X,x\not\in B$ , so it follows that $x\in(X-A)\cap(X-B)$ . Hence $X-(A\cup B)\subset(X-A)\cap(X-B)$ .

Now let $x\in(X-A)\cap(X-B)$ . Then $x\in X-A$ and $x\in X-B$ . By definition of set difference, $x\in X,x\not\in A$ and $x\in X,x\not\in B$ . Since $x\not A,x\not\in B$ , we have $x\not\in(A\cup B)$ , and so $x\in X-(A\cup B)$ . Hence $(X-A)\cap(X-B)\subset X-(A\cup B)$ .

The two sets are subsets of one another, so they must be equal.

$X-\left(\bigcup\limits_{i\in I}A_i\right)=\bigcap\limits_{i\in I}(X-A_i)$

The proof of this is the same as above, you just have to indicate that membership, of lack thereof, holds for all indices $i\in I$ .

Proof of one direction for example:

Let $x\in X-\left(\bigcup\limits_{i\in I}A_i\right)$ . Then $x\in X$ and $x\not\in\bigcup\limits_{i\in I}A_i$ , which in turn means $x\not\in A_i$ for all $i\in I$ . This means $x\in X,x\not\in A_{i_1}$ , and $x\in X,x\not\in A_{i_2}$ , and so on, where $\{i_1,i_2,\ldots\}\subset I$ , for all $i\in I$ . This means $x\in X-A_{i_1}$ , and $x\in X-A_{i_2}$ , and so on, so $x\in\bigcap\limits_{i\in I}(X-A_i)$ . Hence $X-\left(\bigcup\limits_{i\in I}A_i\right)\subset\bigcap\limits_{i\in I}(X-A_i)$ .