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3 years ago

Solve 7x-5x-3=x+4 please help

Mathematics

2 answers:

dybincka [34]3 years ago

5 0

7x-5x-3=x+4
2x-x=4+3
x=7

erastova [34]3 years ago

4 0

2x-3=x+4
x=7
You first do 7x-5x, then you move -3 to the other side, and you move the x to the left side, so it makes it 2x-x=3+4 so x=7

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Syrus is buying a tent with the dimensions shown below. The volume inside the tent is 36 feet. Syrus isn't sure if the tent will

ivolga24 [154]

Answer:

the answer is 3 feet

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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3 years ago

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Please show your work and explain it.

Maru [420]

Answer:

$f(x)=\dfrac{x+2}{2(x-2)}$

Step-by-step explanation:

Remember when you divide fractions, you need to get the reciprocal of the divisor and multiply. So your first simplification would be:

$\dfrac{x^2+4x+4}{x^2-6x+8}\div\dfrac{6x+12}{3x-12}\\\\=\dfrac{x^2+4x+4}{x^2-6x+8}\times\dfrac{3x-12}{6x+12}\\\\=\dfrac{(x^2+4x+4)(3x-12)}{(x^2-6x+8)(6x+12)}$

Next we factor what we can so we can further simplify the rest of the equation:

$=\dfrac{(x^2+4x+4)(3x-12)}{(x^2-6x+8)(6x+12)}\\\\=\dfrac{(x+2)(x+2)(3x-12)}{(x^2-6x+8)(6(x+2))}\\\\$

We can now cancel out (x+2)

$=\dfrac{(x+2)(3x-12)}{(x^2-6x+8)(6)}$

Next we factor out even more:

$=\dfrac{(x+2)(3)(x-4)}{(x-2)(x-4)(6)}$

We cancel out x-4 and reduce the 3 and 6 into simpler terms:

$=\dfrac{(x+2)(1)}{(x-2)(2)}$

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3 years ago

Expand using the properties and rules for logarithms

malfutka [58]

Consider expression $\log_{\frac{1}{2}}\left(\dfrac{3x^2}{2}\right).$

1. Use property

$\log_a\dfrac{b}{c}=\log_ab-\log_ac.$

Then

$\log_{\frac{1}{2}}\left(\dfrac{3x^2}{2}\right)=\log_{\frac{1}{2}}3x^2-\log_{\frac{1}{2}}2.$

2. Use property

$\log_abc=\log_ab+\log_ac.$

Then

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3. Use property

$\log_ab^k=k\log_ab.$

Then

$\log_{\frac{1}{2}}\left(\dfrac{3x^2}{2}\right)=\log_{\frac{1}{2}}3+\log_{\frac{1}{2}}x^2-\log_{\frac{1}{2}}2=\log_{\frac{1}{2}}3+2\log_{\frac{1}{2}}x-\log_{\frac{1}{2}}2.$

4. Use property

$\log_{a^k}b=\dfrac{1}{k}\log_ab.$

Then

$\log_{\frac{1}{2}}\left(\dfrac{3x^2}{2}\right)=\log_{\frac{1}{2}}3+2\log_{\frac{1}{2}}x-\log_{\frac{1}{2}}2=\log_{\frac{1}{2}}3+2\log_{\frac{1}{2}}x-\log_{2^{-1}}2=\\ \\=\log_{\frac{1}{2}}3+2\log_{\frac{1}{2}}x+\log_22=\log_{\frac{1}{2}}3+2\log_{\frac{1}{2}}x+1.$

Answer: correct option is B.

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3 years ago

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Jet001 [13]

Answer: 536 feet tall

Step-by-step explanation:

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3 years ago