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3 years ago

10

17=3(g+3)-g. I also need the math

2 answers:

Feliz [49]3 years ago

4 0

Let's solve your equation step-by-step.<span>17=<span><span>3<span>(<span>g+3</span>)</span></span>−g</span></span>
Step 1: Simplify both sides of the equation.<span>17=<span><span>3<span>(<span>g+3</span>)</span></span>−g</span></span><span>
Simplify: (Show steps)</span><span>17=<span><span>2g</span>+9</span></span>
Step 2: Flip the equation.<span><span><span>2g</span>+9</span>=17</span>
Step 3: Subtract 9 from both sides.<span><span><span><span>2g</span>+9</span>−9</span>=<span>17−9</span></span><span><span>2g</span>=8</span>
Step 4: Divide both sides by 2.<span><span><span>2g</span>2</span>=<span>82</span></span><span>g=4</span>
Answer:<span>g=<span>4</span></span>

IgorC [24]3 years ago

4 0

Do you need it simplified if you do then:

you multiply 3 into the parentheses 17=3(g+3)-g
17=3g+9-g
then add like terms 17=3g+9-g meaning g
17=3g+9

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What is the zero for 2x2-7=4

rosijanka [135]

2x^2-7=4
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3 years ago

Read 2 more answers

The diameter of a wheel of a car is 30cm. If the car travels at an average speed of 13.2m/s, find the number of revolutions made

Marysya12 [62]

Answer:

Approximately $50420$ revolutions (rounded to the nearest whole number.)

Step-by-step explanation:

Convert the diameter of this wheel to meters: $d = 30\; \rm cm = 0.30\; \rm m$ .

Circumference of this wheel:

$\pi \, d = \pi \times 0.30\; \rm m \approx 0.942\; \rm m$ .

Distance travelled in one hour:

$\begin{aligned}v \cdot t &= 13.2 \; \rm m \cdot s^{-1} \times 3600\; \rm s \\ &= 47520\; \rm m\end{aligned}$ .

Number of revolutions required for covering that distance:

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2 years ago

Please help me 33/2 + 3y/5 = 7y/10 + 15

8090 [49]

We are given with an equation in <em>variable y</em> and we need to solve for <em>y</em> . So , now let's start !!!

We are given with ;

${:\implies \quad \sf \dfrac{33}{2}+\dfrac{3y}{5}=\dfrac{7y}{10}+15}$

Take LCM on both sides :

${:\implies \quad \sf \dfrac{165+6y}{10}=\dfrac{7y+150}{10}}$

<em>Multiplying</em> both sides by <em>10</em> ;

${:\implies \quad \sf \cancel{10}\times \dfrac{165+6y}{\cancel{10}}=\cancel{10}\times \dfrac{7y+150}{\cancel{10}}}$

${:\implies \quad \sf 165+6y=7y+150}$

Can be <em>further written</em> as ;

${:\implies \quad \sf 7y+150=165+6y}$

Transposing <em>6y </em>to<em> LHS</em> and <em>150</em> to<em> RHS </em>

${:\implies \quad \sf 7y-6y=165-150}$

${:\implies \quad \bf \therefore \quad \underline{\underline{y=15}}}$

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natita [175]

Answer:

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Step-by-step explanation:

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