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3 years ago

The absolute value of a number is its distance from 0 on a number line. True or False ?

Mathematics

2 answers:

Aleonysh [2.5K]3 years ago

4 0

true.

but the absolute value of 3 is 3 and -3 is 3. 6=6 -6=6

Nina [5.8K]3 years ago

4 0

Answer:

true

Step-by-step explanation:

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Salma is training for a marathon. This week she wants to run a total of 60 miles. She ran 8.55 miles on each of the first 3 days

MArishka [77]

Answer:

51.45 miles

Step-by-step explanation:

7 days = 60 miles

3 days = 8.55 miles

60 - 8.55 = 51.45 miles

6 0

3 years ago

G(n) = n + 3<br> f(n) = n + 1<br> Find (g•f)(-4)

malfutka [58]

He did this to this 64

4 0

3 years ago

Urgent help please!!!!!!!!!!!!!!!!!!

AysviL [449]

Answer:

B. AAS Congruence Theorem

Step-by-step explanation:

Previous steps showed congruence of a side of the designated triangle, and two angles that do not bracket that side. In short form, you have shown congruence of ...

Angle - Angle - Side

so the AAS congruence theorem applies.

8 0

4 years ago

Rationalize the denominator and simplify.

cestrela7 [59]

If you're using the app, try seeing this answer through your browser: brainly.com/question/3151018

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$\mathsf{\dfrac{3\sqrt{6}+5\sqrt{2}}{4\sqrt{6}-3\sqrt{2}}}$

Multiply and divide by the conjugate of the denominator, which is $\mathsf{4\sqrt{6}+3\sqrt{2}:}$

$\mathsf{=\dfrac{\big(3\sqrt{6}+5\sqrt{2}\big)\cdot \big(4\sqrt{6}+3\sqrt{2}\big)}{\big(4\sqrt{6}-3\sqrt{2}\big)\cdot \big(4\sqrt{6}+3\sqrt{2}\big)}}$

Expand those products and eliminate the brackets:

$\mathsf{=\dfrac{\big(3\sqrt{6}+5\sqrt{2}\big)\cdot 4\sqrt{6}+ \big(3\sqrt{6}+5\sqrt{2}\big)\cdot 3\sqrt{2}}{\big(4\sqrt{6}-3\sqrt{2}\big)\cdot 4\sqrt{6}+ \big(4\sqrt{6}-3\sqrt{2}\big)\cdot 3\sqrt{2}}}\\\\\\ \mathsf{=\dfrac{12\cdot \big(\sqrt{6}\big)^2+20\sqrt{2}\cdot \sqrt{6}+9\cdot \sqrt{6}\cdot \sqrt{2}+15\cdot \big(\sqrt{2}\big)^2}{16\cdot \big(\sqrt{6}\big)^2-12\cdot \sqrt{2}\cdot \sqrt{6}+ 12\cdot \sqrt{6}\cdot \sqrt{2}-9\cdot \big(\sqrt{2}\big)^2}}\\\\\\ \mathsf{=\dfrac{12\cdot 6+20\sqrt{2\cdot 6}+9\sqrt{6\cdot 2}+15\cdot 2}{16\cdot 6-12\sqrt{2\cdot 6}+ 12\sqrt{6\cdot 2}-9\cdot 2}}\\\\\\ \mathsf{=\dfrac{72+20\sqrt{12}+9\sqrt{12}+30}{96-12\sqrt{12}+12\sqrt{12}-18}}$

$\mathsf{=\dfrac{72+29\sqrt{12}+30}{96-18}}\\\\\\ \mathsf{=\dfrac{102+29\sqrt{2^2\cdot 3}}{78}}\\\\\\ \mathsf{=\dfrac{102+29\cdot \sqrt{2^2}\cdot \sqrt{3}}{78}}\\\\\\ \mathsf{=\dfrac{102+29\cdot 2\sqrt{3}}{78}}$

$\mathsf{=\dfrac{102+58\sqrt{3}}{78}}\\\\\\ \mathsf{=\dfrac{\,\diagup\!\!\!\! 2\cdot \big(51+29\sqrt{3}\big)}{\diagup\!\!\!\! 2\cdot 39}}$

$\mathsf{=\dfrac{51+29\sqrt{3}}{39}}\quad\longleftarrow\quad\textsf{this is the answer.}$

I hope this helps. =)

4 0

3 years ago

Somebody help me so I can give y’all some points

Norma-Jean [14]

Answer:

x=1

Step-by-step explanation:

3 0

2 years ago