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3 years ago

Solve. d + 4 = –1 A. –5 B. –3 C. 3 D. 5

Mathematics

2 answers:

Marina86 [1]3 years ago

8 0

The answer is A. -5 there is a site called Symbolab that solves them for you.

77julia77 [94]3 years ago

7 0

The answer to your question is -5

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Can someone please answer this?

galina1969 [7]

Since we know the side length of the square (6), we can calculate its diagonal using pythagoras.

diag d = √(6²+6²) = 6√2 in

The diagonal is also the diameter of the circle! So the radius of the circle is half of that:

radius r = d/2 = 3√2 in

The area of the circle is πr² = π(3√2)² = 18π in²

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3 years ago

What is the number and the question of 2+2

Maurinko [17]

The answer is 4. The question is what is 2+2?

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3 years ago

A horse travels an average speed of 12 miles per hour how long does it take a horse to travel 60

olganol [36]

5 hours hun just divide and times and maybe you will find it

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3 years ago

Ill give Brainliest. please explain your answers. im so confused on these questions :((

Katen [24]

Answer:

I'll answer the first one. Neither vehicles are crossing the speed limit since both are under 40mph.

Step-by-step explanation:

6 miles every 10 minutes = 36 miles every hour.

15 miles every 25 minutes = 36 miles every hour.

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Neither vehicles are crossing the speed limit since both are under 40mph.

Hope this helped! :))

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

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3 years ago