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2 years ago

Are constants like terms?

Mathematics

1 answer:

aniked [119]2 years ago

7 0

Answer:

yes they are

Step-by-step explanation:

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What is 20/-12 as a mixed number

kompoz [17]

if we divide 20 by 12, we get a quotient of 1, and a remainder of 8.

$\bf -\cfrac{20}{12}\implies -\stackrel{quotient}{\text{\huge 1}}\frac{\stackrel{remainder}{8}}{12}\implies -1\frac{8}{12}$

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3 years ago

Please I just need help with these two problems

Oksana_A [137]

$3d-4+d=8d+12 \implies \\4d-4=8d+12 \implies-4d=16 \implies\\ d=-4$

$f-6=-2f+3(f-2) \implies \\ f-6=-2f+3f-6 \implies\\ f-6=f-6 \implies\\ 0=0 \implies \text{ all real numbers are a solution to this problem}$

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2 years ago

Read 2 more answers

Is my answer right? the other options are A.0.33 B.0.42 C.0.58 D.0.67

Fittoniya [83]

Yes 0.42 is correct!!

4 0

3 years ago

Nina and Cindy went to the early show. Tickets were $3.00 each. They shared a jumbo popcorn for $4.50 and each girl bought a sma

AlexFokin [52]

Answer:

$18.50

Step-by-step explanation:

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3 years ago

Read 2 more answers

13.......................

Alla [95]

Answer:

The answer is (d) ⇒ $pq^{2}r\sqrt[3]{pr^{2}}$

Step-by-step explanation:

* To simplify the cube roots:

If its number then the number must be written in the form x³

then we divide the power by 3 to cancel the radical

If its variable we divide its power by 3 to cancel the radical

∵ $\sqrt[3]{p^{4}q^{6}r^{5}}=p^{\frac{4}{3}}q^{\frac{6}{3}}r^{\frac{5}{3}}}$

∴ $p^{\frac{4}{3}}q^{2}}r^{\frac{5}{3}}=p^{1\frac{1}{3}}q^{2}r^{1\frac{2}{3}}$

∵ $p^{\frac{1}{3}}=\sqrt[3]{p}$

∵ $r^{\frac{2}{3}}=\sqrt[3]{r^{2}}$

∴ $p(p)^{\frac{1}{3}}q^{2}r(r)^{\frac{2}{3}}=p(\sqrt[3]{p})q^{2}r(\sqrt[3]{r^{2}})$

∴ $prq^{2}\sqrt[3]{pr^{2}}}$

∴ The answer is (d)

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2 years ago