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Nikolay [14]
3 years ago
13

What is the absolute value of -3

Mathematics
1 answer:
scoray [572]3 years ago
6 0
3
it’s just how far away that number is from zero
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How many ways can 4 pizza toppings be chosen from 12 available toppings
Arisa [49]

Answer:

There are 495 different ways to choose 4 pizza toppings from 12.

Step-by-step explanation:

12!/8!4! = 12*11*10*9/(4*3*2) = 11*9*5 = 495

5 0
3 years ago
Read 2 more answers
What's 6/7 written as a fraction with a denominator of 42? A. 36/42 b. 7/42 c. 42/42 d. 6/42
7nadin3 [17]

Pls mark Brainliest.

Answer:

A. 36/42

Step-by-step explanation:

What do you multiply 7 by to get 42?

6. That is what the denominator was multiplied by, and so, the equivalent fraction must have a numerator that is 6 times larger than the original.

6 is the numerator so the numerator of the equivalent fraction with denominator 42 must be 36 since 6 * 6 = 36/

A. 36/42

4 0
3 years ago
Describe 58 as a sum of tens and ones.
d1i1m1o1n [39]
50+8 50 represents tens place value, while 8 represents the ones place value.
6 0
3 years ago
Find the answer to zero point 73 repeating times square root of 147 end square root and select the correct answer below. Also, i
liubo4ka [24]

Answer:

\frac{511\,\,\sqrt{3} }{99}  which is an irrational number

Step-by-step explanation:

Recall that the repeating decimal 0.7373737373... can be written in fraction form as: \frac{73}{99}

Now, let's write the number 147 which is inside the square root in factor form to find if it has some perfect square factors:

147=7^2\,3

Then, 7 will be able to go outside the root when we compute the final product requested:

\frac{73}{99} \,*\,7\,\sqrt{3} =\frac{511\,\,\sqrt{3} }{99}

This is an irrational number due to the fact that it is the product of a rational number (quotient between 511 and 99) times the irrational number \sqrt{3}

5 0
3 years ago
(1) Let {v1,v2,v3} be a set of vectors in Rn . If u is Span {v1,v2,v3}, show that 3u is in Span {v1,v2,v3}.
Evgesh-ka [11]

Answer:

(1)

Multiplying by 3 both sides of the equality you get that

3u = 3c_1v_1+3c_2v_2+3c_3v_3

3u  is in the Span of the vectors \{v_1,v_2,v_3\}.

(2)

That's not true, consider the following counter example.

v_1 = (0,0,0,1)\\v_2 = (0,0,1,0)\\v_3 = (0,1,0,0)\\v_4 = (1,0,0,0)\\u = (0,1,1,1)

u is a linear combination of v_1,v_2,v_3 but is NOT a linear combination of v_1,v_2,v_3,v_4.

Step-by-step explanation:

(1)

As the hint indicates, you know that

u = c_1 v_1 + c_2v_2+c_3v_3

Then, if you multiply both sides of the equality by 3, you get that

3u = 3c_1v_1+3c_2v_2+3c_3v_3

And that's it. 3u  is in the Span of the vectors \{v_1,v_2,v_3\}

(2)

That's not true, consider the following counter example.

v_1 = (0,0,0,1)\\v_2 = (0,0,1,0)\\v_3 = (0,1,0,0)\\v_4 = (1,0,0,0)\\u = (0,1,1,1)

u is a linear combination of v_1,v_2,v_3 but is NOT a linear combination of v_1,v_2,v_3,v_4.

4 0
3 years ago
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