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2 years ago

Help this is due today?

Mathematics

2 answers:

just olya [345]2 years ago

8 0

Answer:

2s for the first box

6u for the second box

Step-by-step explanation:

when they say like terms, just think about how the letter parts of the expression are the same

like 1/2s and 2s both have the singular variable s

and how 6u and 9u both have the singular variable u

Mashutka [201]2 years ago

3 0

To solve this one you have to find the ones that are connected by variables or look the same

For example.

1x and 2x are like terms because they have the same variable X

4d and 6/2d are like terms because they have the same variable D

4 and 3 are like terms because they have no variables, and they are the same

So now that you understand what like terms are you may so it

1/2 and 2s are like terms because they share the same variable S

6u and 9u are like terms because they share the same variable U

I hope this helped and if it's wrong blame me

Ps. Brainlist would really help

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(6b+9)–(6b+5) . . . . . . . .

fiasKO [112]

Answer:

4

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Please help!posted picture of question

ale4655 [162]

We have the following arithmetic series:
4, 13, 22, 31, 40
The common difference is given by:
13-4 = 9
22-13 = 9
31-22 = 9
40-31 = 9
Since it is an arithmetic series, the difference is constant.
Answer:
The common difference is:
9
option 4

7 0

3 years ago

Simplify the expression. Assume that all variables represent nonzero real numbers.StartFraction (4 n Superscript 4 Baseline q Su

Alja [10]

Answer:

$\frac{ - 3}{ 256 {q}^{10} {n}^{8} }$

Step by step explanation:

$\frac{ {(4 {n}^{4} {q}^{5})}^{2} {(8 {n}^{4} q)}^{-2} }{ {(- 3 {nq}^{9})}^{ - 1} {(4 {n}^{3} {q}^{9}) }^{3} }$

first we will change the terms with negative superscrips to the other side of the fraction

$\frac{{(4 {n}^{4} {q}^{5})}^{2}{(- 3 {nq}^{9})}^{ 1}}{{(4 {n}^{3} {q}^{9})}^{3} {(8 {n}^{4} q)}^{2} }$

then we will distribute the superscripts

$\frac{ {4}^{2} {n}^{2 \times 4} {q}^{2 \times 5} (- 3) {nq}^{9}}{ {4 }^{3}{n}^{3 \times 3} {q}^{9 \times 3} {8 }^{2}{n}^{4 \times 2} {q}^{2} }$

$\frac{ {4}^{2} {n}^{8} {q}^{10} (- 3) {nq}^{9}}{ {4 }^{3}{n}^{9} {q}^{27} {8 }^{2}{n}^{8} {q}^{2} }$

as when multiplying two powers that have the same base, we can add the exponents and, to divide podes with the same base, we can subtract the exponents

${4}^{2 - 3} {q}^{10 + 9 - 2 - 27} {n}^{8 + 1 - 8 - 9} {8}^{ - 2} { (- 3)}^{1}$