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enot [183]
2 years ago
10

Solve each equation. 2p = 2 p = _____q - 3 = 7 q = _____​

Mathematics
2 answers:
svp [43]2 years ago
8 0
P=1
q= 10
p=1 because 2x1 equals 2
q= 10, because 10 minus 3=7(just add 7+3)
Lapatulllka [165]2 years ago
3 0

Answer:

\textbf{HELLO!!}

2p=2

\frac{2p}{2}=\frac{2}{2} \hookleftarrow \mathrm{Divide\:both\:sides\:by\:}2

=1

\boxed{\boxed{\underline{\textsf{\textbf{P=1}}}}}

---------------------

q - 3 = 7

q-3+3=7+3 \hookleftarrow \mathrm{Add\:}3\mathrm{\:to\:both\:sides}

=10

\boxed{\boxed{\underline{\textsf{\textbf{q=10}}}}}

-----------------------

\textbf{HOPE IT HELPS}

\textbf{HAVE A GREAT DAY!!}

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bulgar [2K]
2/6 = 1/3
2/5 = 4/10 

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x \geqslant ( - 32) \:  \:  \: \:  or \:  \:  \:  \: ( - 32) \leqslant x

Step-by-step explanation:

\frac{x}{4}  \geqslant  - 8 \\ x \geqslant  - 8 \times 4 \\ x \geqslant  (- 32) \\

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Subtract 1/2h+1 from 3/4 h+4
damaskus [11]
-h/4+5 is your answer



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8 0
3 years ago
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Kiran scored 36 baskets in 1 week. She scored 1/2 of the baskets on the weekend. She scored 1/4 of them on Tuesday, she scored 1
notka56 [123]

Answer:

6 baskets did Kiran score on Thursday

Step-by-step explanation:

As per the statement:

Kiran scored 36 baskets in 1 week.

On weekend she scored = \frac{1}{2} \cdot 36 = 18 baskets.

She scored 1/4 of them on Tuesday

⇒On Tuesday she scored = \frac{1}{4} \cdot 36 = 9 baskets

She scored 1/12 of them on Wednesday

⇒On Wednesday she scored = \frac{1}{12} \cdot 36 = 3 baskets.

Then;

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Therefore, 6 baskets did Kiran score on Thursday

8 0
3 years ago
Can someone help me in this plssss I really need it
kompoz [17]

Answer:

\boxed{-3xy^{2}\sqrt [3] {2x^{2}}}

Step-by-step explanation:

Your expression is

\sqrt [3] {-54x^{5}y^{6}}

Here's how I would simplify it.

\begin{array}{rcll}\sqrt [3] {-54x^{5}y^{6}} & = & \sqrt [3] {(-1)^{3}\times 2 \times 27 \times x^{2} \times x^{3} \times y^{6}} & \text{Factored the cubes}\\& = & \sqrt [3] {(-1)^{3} \times 3^{3}\times x^{3} \times y^{6}\times 2 \times x^{2}} & \text{Grouped the cubes}\\\end{array}

\begin{array}{rcll}& = & \sqrt [3] {(-1)^{3} \times {3^{3}\times x^{3} \times y^{6}}} \times\sqrt [3] { 2 \times x^{2}} & \text{Separated the cubes}\\&=& \mathbf{-3xy^{2}\sqrt [3] {2x^{2}}} & \text{Took cube roots}\\\end{array}

\text{The simplified expression is $\boxed{\mathbf{-3xy^{2}\sqrt [3] {2x^{2}}}}$}

6 0
3 years ago
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