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3 years ago

What is the absolute value of –38? A. 19 B. –19 C. 38 D. –38

Mathematics

1 answer:

GREYUIT [131]3 years ago

6 0

<u>Answer</u>

C. 38

<u>Explanation</u>

The absolute value of any digit is the magnitude or its size. The negative side is not considered.

It can also be defined as the actual magnitude of a numerical value or measurement, irrespective of its relation to other values.

It is represented as; /x/.

The absolute value of -38 is /-38/.

∴ /-38/ = 38

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Can u please do 1,2

Sidana [21]

1) First, you have to add all the 3 pieces of wood
10+7.5+12= 29.5 ft.
Second, you have to add the pieces that he has used
5.2 +6.2= 11.4 ft
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29.5 - 11.4=18.1 ft.
I hope it helped. But, sorry don't know how to do number 2.

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3 years ago

A six-sided dice is rolled twice. Find the probability that the larger of the two rolls was equal to 5.

Diano4ka-milaya [45]

The probability that the larger of the two rolls was equal to 5 is; 2/9

<h3>Probability of possible outcomes</h3>

The total number of possible outcomes when the dice is rolled twice is 36 outcomes.

Now, for the larger of the two rolls to equal 5, the total number of favourable outcomes are 8 which are;

(1, 5), (2, 5), (3, 5), (4, 5), (5, 1), (5, 2), (5, 3), (5,4)

Thus, probability that the larger of the two rolls was equal to 5 is;

P(larger of 2 rolls equals 5) = 8/36 = 2/9

Read more about probability of possible outcomes at; brainly.com/question/19916581

8 0

2 years ago

Help me solve -2+5n=4n+2

iris [78.8K]

Sorry for the messy handwriting

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3 years ago

Please Help!!

Genrish500 [490]

Given

$a\sqrt{x+b}+c=d$

we have

$\sqrt{x+b}=\dfrac{d-c}{a}$

Squaring both sides, we have

$x+b=\dfrac{(d-c)^2}{a^2}$

And finally

$x=\dfrac{(d-c)^2}{a^2}-b$

Note that, when we square both sides, we have to assume that

$\dfrac{d-c}{a}>0$

because we're assuming that this fraction equals a square root, which is positive.

So, if that fraction is positive you'll actually have roots: choose

$a=1,\ b=0,\ c=2,\ d=6$

and you'll have

$\sqrt{x}+2=6 \iff \sqrt{x}=4 \iff x=16$

Which is a valid solution. If, instead, the fraction is negative, you'll have extraneous roots: choose

$a=1,\ b=0,\ c=10,\ d=4$

and you'll have

$\sqrt{x}+10=4 \iff \sqrt{x}=-6$

Squaring both sides (and here's the mistake!!) you'd have

$x=36$

which is not a solution for the equation, if we plug it in we have

$\sqrt{x}+10=4 \implies \sqrt{36}+10=4 \implies 6+10=4$

Which is clearly false

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3 years ago

Suppose you know three minimum,lower quartile,median, upper quartile, and maximum values for a set of data. Explain how to creat

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3 years ago