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3 years ago

Can someone help me answer this please!!!

Mathematics

1 answer:

galben [10]3 years ago

3 0

Hello!

Your answer is...

14x/x-2 - 28/x-2 = ==> 14

- Dukuccino<span>©</span> ❤

Hope this helps! ^^ Brainlyist please

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Hurt just hurry and answer please

Luden [163]

Answer:

Step-by-step explanation:

Solve:

6*7= 42

42*2.00=84

Hope this helps!

3 0

3 years ago

A rectangular garden has length and width as given by the expressions below.

Tom [10]

Answer:

P = -12xy - 56y - 42x + 8

Step-by-step explanation:

P = 2L + 2W

P = 2(4 - 7(3x + 4y)) + 2(3x(-2y))

P = 8 - 14(3x + 4y) + 6x(-2y)

P = 8 - 42x - 56y - 12xy

P = -12xy - 56y - 42x + 8

3 0

3 years ago

Read 2 more answers

A bottle of vitamins should have 60 vitamins. The actual number is 62. What is the percent error to the nearest hundredth?

Phoenix [80]

Answer:

3.33%

Step-by-step explanation:

(62/60)X100=103.33 meaning the bottle was overfilled by 3.33%

Brainliest please

8 0

3 years ago

Read 2 more answers

Tell whether the table of values represents a

spin [16.1K]

Exponential Function.

Plotting the graph using this table gives the first figure. An exponential function has this kind of graph.

4 0

2 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

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

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.

7 0

3 years ago