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Virty [35]
3 years ago
15

Solve for x. i will leave pictuere below

Mathematics
2 answers:
Andreyy893 years ago
8 0

3/2x=3/4 (cross multiplied)

X=1/2 (divided both sides by 3/2)

Veronika [31]3 years ago
4 0

\dfrac{x}{3}=\dfrac{\dfrac{1}{4}}{\dfrac{3}{2}}\\\\ \dfrac{x}{3}=\dfrac{1}{6}\\\\ 6x=3\\\\ x=\dfrac{3}{6}=\dfrac{1}{2}

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Answer:

x=-27, y=-15

Step-by-step explanation:

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3 years ago
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The perimeter is 36 and the area is 81 units
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4 over 5 is the correct answer
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The radioactive element​ carbon-14 has a​ half-life of 5750 years. A scientist determined that the bones from a mastodon had los
Gnesinka [82]

The bones at the time they were​ discovered when the radioactive element​ carbon-14 has a​ half-life of total 5750 years are 10062.5-year-old.

<h3>What is half-lives?</h3>

Half lives is the time interval which is need to decay the atomic nuclei of a radioactive sample.

There is a scientist who determined that the bones from a mastodon had lost 70.3​% of their​ carbon-14.  Thus, the fraction remaining is,

f=1-(70.3/100)=1-0.703

f=1-(70.3/100)=0.297

Now the fraction remaining can be given as,

f=(1/2)ⁿ

Here, n is the half life elapsed. Put the value of fraction remaining.

0.297=(1/2)ⁿ

n=1.75

The radioactive element​ carbon-14 has a​ half-life of total 5750 years. Thus,

Years=1.75*5750

Years=10062.5

Thus, the bones at the time they were​ discovered when the radioactive element​ carbon-14 has a​ half-life of total 5750 years are 10062.5-year-old.

Learn more about the half lives here;

brainly.com/question/2320811

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6 0
2 years ago
[tex]cos {}^{4} α+sin {}^{4} α= \frac{1}{4} (3+cos4α)<br>Prove:<br>​
asambeis [7]

Given:

\cos^4 \alpha+\sin^4\alpha=\dfrac{1}{4}(3+\cos 4 \alpha)

To prove:

The given statement.

Proof:

We have,

\cos^4 \alpha+\sin^4\alpha=\dfrac{1}{4}(3+\cos 4 \alpha)

LHS=\cos^4 \alpha+\sin^4\alpha

LHS=(\cos^2 \alpha)^2+(\sin^2 \alpha)^2

LHS=(\cos^2 \alpha+\sin^2\alpha)^2-2\sin ^2\alpha\cos^2 \alpha     [\because a^2+b^2=(a+b)^2-2ab]

LHS=(1)^2-2(1-\cos^2 \alpha)\cos^2 \alpha      [\because \cos^2 \alpha+\sin^2\alpha=1]

LHS=1-2\cos^2 \alpha+2\cos^4 \alpha

Now,

RHS=\dfrac{1}{4}(3+\cos 4 \alpha)

RHS=\dfrac{1}{4}[3+(2\cos^2 2\alpha-1)]        [\because \cos 2\theta=2\cos^2\theta -1]

RHS=\dfrac{1}{4}[2+2\cos^2 2\alpha]

RHS=\dfrac{1}{4}[2+2(2\cos^2 \alpha-1)^2]        [\because \cos 2\theta=2\cos^2\theta -1]

RHS=\dfrac{1}{4}[2+2(4\cos^4 \alpha-4\cos \alpha+1)]        [\because (a-b)^2=a^2-2ab+b^2]

RHS=\dfrac{1}{4}[2+8\cos^4 \alpha-8\cos \alpha+2]

RHS=\dfrac{1}{4}[4+8\cos^4 \alpha-8\cos \alpha]

RHS=1+2\cos^4 \alpha-2\cos \alpha

RHS=1-2\cos^2 \alpha+2\cos^4 \alpha

LHS=RHS

Hence proved.

8 0
3 years ago
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