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Mariana [72]
4 years ago
5

Im sorry, im not good with negatives and positives xD this is the last one Which values satisfy the inequality? |x| < 3 Choos

e all answers that are correct.
a. x = –6
b. x = –5
c. x = –1
d. x = 2
Mathematics
1 answer:
Katarina [22]4 years ago
8 0

No.  You can't squirm out of it that easily !  Sorry.

The absolute value of a number is its size without a plus or minus.
So you don't need to be good with plus and minus to work absolute values.

The statement says "the absolute value of 'x' is less than 3".
Let's test each choice and see if it works:

a). x = -6.  The absolute value of  -6  is  6.  Is that less than  3 ?  No.

b). x = -5.  The absolute value of  -5  is  5.  Is that less then  3 ?  No.

c). x = -1.  The absolute value of  -1  is  1.  Is that less than  3 ?  Yes.

d). x = 2.  The absolute value of  2  is  2. Is that less than  3 ?  Yes.

Do you see ?  You don't need to be Einstein or a math whiz
to answer this one.

You're just giving up because people say that girls can't do math,
and instead of telling them where to stick it, you bought into it.

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Answer:

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Step-by-step explanation:

The quadrilateral ABCD consists of two triangles. By adding the area of the two triangles, we get the area of the entire quadrilateral.

Vertices A, B, and C form a right triangle with legs AB=3, BC=4, and AC=5. The two legs, 3 and 4, represent the triangle's height and base, respectively.

The area of a triangle with base b and height h is given by A=\frac{1}{2}bh. Therefore, the area of this right triangle is:

A=\frac{1}{2}\cdot 3\cdot 4=\frac{1}{2}\cdot 12=6\:\mathrm{cm^2}

The other triangle is a bit trickier. Triangle \triangle ADC is an isosceles triangles with sides 5, 5, and 4. To find its area, we can use Heron's Formula, given by:

A=\sqrt{s(s-a)(s-b)(s-c)}, where a, b, and c are three sides of the triangle and s is the semi-perimeter (s=\frac{a+b+c}{2}).

The semi-perimeter, s, is:

s=\frac{5+5+4}{2}=\frac{14}{2}=7

Therefore, the area of the isosceles triangle is:

A=\sqrt{7(7-5)(7-5)(7-4)},\\A=\sqrt{7\cdot 2\cdot 2\cdot 3},\\A=\sqrt{84}, \\A=2\sqrt{21}\:\mathrm{cm^2}

Thus, the area of the quadrilateral is:

6\:\mathrm{cm^2}+2\sqrt{21}\:\mathrm{cm^2}=\boxed{6+2\sqrt{21}\:\mathrm{cm^2}}

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Answer:

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Step-by-step explanation:

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