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3 years ago

Can someone please help me I need it ASAP

Mathematics

1 answer:

VLD [36.1K]3 years ago

5 0

We're not supposed to help with exams. This isn't an exam, is it?

A(0)=202, A(37)=half that.

$A(t) = 202 (\frac 1 2)^{t/37}$

$A(100) = 202 (\frac 1 2)^{100/37} \approx 31.0$

Last choice.

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PLLLLLLSSSSSs helpppppppp

Citrus2011 [14]

Answer:

x = 40

y = 62

Step-by-step explanation:

I figured out the answer by finding the x on the left triangle first by writing an equation:

(3x - 18) + (x + 7) + (x - 9) = 180

once I worked that I plugged in the number which was 40.

Then for the triangle to the right I also plugged in the numbers for x and got 118 degrees I took 180 and subtracted 118 to get my answer of 62 for y.

5 0

3 years ago

The probability that a can of paint contains contamination is 3.23%, and the probability of a mixing error is 2.4%. The probabil

stich3 [128]

Answer:

4.6%.

Step-by-step explanation:

The probability that a can of paint contains contamination(C) is 3.23%

P(C)=3.23%

The probability of a mixing(M) error is 2.4%.

P(M)=2.4%

The probability of both is 1.03%.

$P(C \cap M)=1.03\%$

We want to determine the probability that a randomly selected can has contamination or a mixing error. i.e. $P(C \cup M)$

In probability theory:

$P(C \cup M) = P(C)+P(M)-P(C \cap M)\\P(C \cup M)=3.23+2.4-1.03\\P(C \cup M)=4.6\%$

The probability that a randomly selected can has contamination or a mixing error is 4.6%.

3 0

3 years ago

which of the following is equivalent to 3 sqrt 32x^3y^6 / 3 sqrt 2x^9y^2 where x is greater than or equal to 0 and y is greater

Nutka1998 [239]

Answer:

$\frac{\sqrt[3]{16y^4}}{x^2}$

Step-by-step explanation:

The options are missing; However, I'll simplify the given expression.

Given

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$

Required

Write Equivalent Expression

To solve this expression, we'll make use of laws of indices throughout.

From laws of indices $\sqrt[n]{a} = a^{\frac{1}{n}}$

So,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ gives

$\frac{(32x^3y^6)^{\frac{1}{3}}}{(2x^9y^2)^\frac{1}{3}}$

Also from laws of indices

$(ab)^n = a^nb^n$

So, the above expression can be further simplified to

$\frac{(32^\frac{1}{3}x^{3*\frac{1}{3}}y^{6*\frac{1}{3}})}{(2^\frac{1}{3}x^{9*\frac{1}{3}}y^{2*\frac{1}{3}})}$

Multiply the exponents gives

$\frac{(32^\frac{1}{3}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

Substitute $2^5$ for 32

$\frac{(2^{5*\frac{1}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

From laws of indices

$\frac{a^m}{a^n} = a^{m-n}$

This law can be applied to the expression above;

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$ becomes

$2^{\frac{5}{3}-\frac{1}{3}}x^{1-3}*y^{2-\frac{2}{3}}$

Solve exponents

$2^{\frac{5-1}{3}}*x^{-2}*y^{\frac{6-2}{3}}$

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$

From laws of indices,

$a^{-n} = \frac{1}{a^n}$ ; So,

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$ gives

$\frac{2^{\frac{4}{3}}*y^{\frac{4}{3}}}{x^2}$

The expression at the numerator can be combined to give

$\frac{(2y)^{\frac{4}{3}}}{x^2}$

Lastly, From laws of indices,

$a^{\frac{m}{n} = \sqrt[n]{a^m}$ ; So,

$\frac{(2y)^{\frac{4}{3}}}{x^2}$ becomes

$\frac{\sqrt[3]{(2y)}^{4}}{x^2}$

$\frac{\sqrt[3]{16y^4}}{x^2}$

Hence,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ is equivalent to $\frac{\sqrt[3]{16y^4}}{x^2}$

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Alla [95]

Answer:

Step-by-step explanation:

It's 110. Since the one leg is not on zero it's not a full 160 instead you have to subtract 50 where the leg is.

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2 years ago

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