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aalyn [17]
3 years ago
10

Find the value of x. m2 = x + 67

Mathematics
2 answers:
FrozenT [24]3 years ago
8 0

Answer:

x=m2-67

Step-by-step explanation:

nydimaria [60]3 years ago
3 0

Answer:

x=m2-67

is the actual answer I

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Can you help me please
Setler [38]
I think this question is a part of another question; you can have any ratio that can add up to 114 cm.
5 0
4 years ago
4. What are the values of the variables in the triangle below? If your answer is not an integer, leave
pentagon [3]

Using relations in a right triangle, it is found that the values of the variables in the triangle below are given by:

x = 18, y = 6\sqrt{3}

<h3>What are the relations in a right triangle?</h3>

The relations in a right triangle are given as follows:

  • The sine of an angle is given by the length of the opposite side to the angle divided by the length of the hypotenuse.
  • The cosine of an angle is given by the length of the adjacent side to the angle divided by the length of the hypotenuse.
  • The tangent of an angle is given by the length of the opposite side to the angle divided by the length of the adjacent side to the angle.

In this problem, the hypotenuse is of 12\sqrt{3}, while the opposite angle to side y is of 30º, hence:

\sin{30^\circ} = \frac{y}{12\sqrt{3}}

\frac{1}{2} = \frac{y}{12\sqrt{3}}

y = 6\sqrt{3}

The adjacent side to the angle of 30º is of x, hence:

\cos{30^\circ} = \frac{x}{12\sqrt{3}}

\frac{\sqrt{3}}{2} = \frac{x}{12\sqrt{3}}

2x = 36

x = 18

More can be learned about relations in a right triangle at brainly.com/question/26396675

7 0
2 years ago
Is the expression 3x+2 a binomial?
Wewaii [24]

Yes, this is a polynomial, but a special form of a polynomial with just one term which is called a monomial .

7 0
3 years ago
Differentiate x^2 e^x logx
zimovet [89]

Product rule:

\dfrac{\mathrm d}{\mathrm dx}(x^2e^x\log x)

=\dfrac{\mathrm d(x^2)}{\mathrm dx}e^x\log x+x^2\dfrac{\mathrm d(e^x)}{\mathrm dx}\log x+x^2e^x\dfrac{\mathrm d(\log x)}{\mathrm dx}

Power rule:

\dfrac{\mathrm d(x^2)}{\mathrm dx}=2x

The exponential function is its own derivative:

\dfrac{\mathrm d(e^x)}{\mathrm dx}=e^x

Assuming the base of \log x is e, its derivative is

\dfrac{\mathrm d(\log x)}{\mathrm dx}=\dfrac1x

But if you mean a logarithm of arbitrary base b, we have

y=\log_bx\implies x=b^y=e^{y\ln b}\implies1=e^{y\ln b}\ln b\dfrac{\mathrm dy}{\mathrm dx}

\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{e^{-y\ln b}}{\ln b}=\dfrac1{b^y\ln b}

\implies\dfrac{\mathrm d(\log_bx)}{\mathrm dx}=\dfrac1{x\ln b}

So we end up with

2xe^x\log x+x^2e^x\log x+\dfrac{x^2e^x}x

=xe^x(2\log x+x\log x+1)

8 0
3 years ago
Can you help me please
sasho [114]
Maybe try to take another pic of the paper so I can see the rest of the question. Then I can try to figure it out. :)
8 0
4 years ago
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