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2 years ago

Do the ratios 6/1 and 18/3 forma proportion yes or no?

Mathematics

1 answer:

andrezito [222]2 years ago

4 0

I'm pretty sure it's no...

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3 years ago

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2(x + 3)=x - 4<br>help me

Irina18 [472]

Answer:

x = -10

Step-by-step explanation:

2(x+3) = x-4 --> multiply (x+2) times 2

2x+6 = x - 4 --> subtract x from both sides

x+6 = -4 --> subtract 6 from both sides

x = -10

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2 years ago

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The school water fountain can only fill 3/4 of a jug at a time. With this amount of water, Mr. Kern can water 3/8 of his plants.

Vesnalui [34]

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$\frac{3}{4} = \frac{3}{3} \\ \frac{3}{8} = x \\ x = \frac{3}{8} \times \frac{4}{3} \\ x = \frac{1}{2}$

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2 years ago

36m⁴-√16 pls help asap

shusha [124]

Answer:

Step-by-step explanation:

$(a^2-b^2)=(a-b)(a+b)\\ \\ 36m^4-4\\ \\ (6m^2-2)(6m^2+2)\\ \\ 4(3m^2-1)(3m^2+1)$

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3 years ago

Help! If you know this can you tell me how to do it?

aleksandr82 [10.1K]

Answer:

Step-by-step explanation:

Here's how this works:

Get everything together into one fraction by finding the LCD and doing the math. The LCD is sin(x) cos(x). Multiplying that in to each term looks like this:

$[sin(x)cos(x)]\frac{sin(x)}{cos(x)}+[sin(x)cos(x)]\frac{cos(x)}{sin(x)} =?$

In the first term, the cos(x)'s cancel out, and in the second term the sin(x)'s cancel out, leaving:

$\frac{sin^2(x)}{sin(x)cos(x)}+\frac{cos^2(x)}{sin(x)cos(x)}=?$

Put everything over the common denominator now:

$\frac{sin^2(x)+cos^2(x)}{sin(x)cos(x)}=?$

Since $sin^2(x)+cos^2(x)=1$ , we will make that substitution:

$\frac{1}{sin(x)cos(x)}$

We could separate that fraction into 2:

$\frac{1}{sin(x)}$ × $\frac{1}{cos(x)}$

$\frac{1}{sin(x)}=csc(x)$ and $\frac{1}{cos(x)}=sec(x)$

Therefore, the simplification is

sec(x)csc(x)

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3 years ago