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Arturiano [62]
3 years ago
8

I do not know how to solve this please help

Mathematics
1 answer:
oksano4ka [1.4K]3 years ago
8 0
You multiply the denominator on one side of the equation by the numerator and vice versa. So for example(S)I'm going to multiply 4 by nine and 3 times n. Then that comes out to 3n=36. After you multiply, you divide from both sides to get the variable by itself. So you would divide both sides by 3 and you're final answer would be n=13. Hope this helped!
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X+y=20 3x-3y=30 Which values of x and y satisfy this system of equations?
puteri [66]

x +y = 20 ( rewrite as x = 20-y)

3x - 3y =30

3(20-y) - 3y = 30

60-3y - 3y = 30

60-6y = 30

-6y = -30

y = -30 / -6 = 5

x=20-5 = 15

x = 15, y = 5

4 0
4 years ago
How to differentiate ?
Bas_tet [7]

Use the power, product, and chain rules:

y = x^2 (3x-1)^3

• product rule

\dfrac{\mathrm dy}{\mathrm dx} = \dfrac{\mathrm d(x^2)}{\mathrm dx}\times(3x-1)^3 + x^2\times\dfrac{\mathrm d(3x-1)^3}{\mathrm dx}

• power rule for the first term, and power/chain rules for the second term:

\dfrac{\mathrm dy}{\mathrm dx} = 2x\times(3x-1)^3 + x^2\times3(x-1)^2\times\dfrac{\mathrm d(3x-1)}{\mathrm dx}

• power rule

\dfrac{\mathrm dy}{\mathrm dx} = 2x\times(3x-1)^3 + x^2\times3(3x-1)^2\times3

Now simplify.

\dfrac{\mathrm dy}{\mathrm dx} = 2x(3x-1)^3 + 9x^2(3x-1)^2 \\\\ \dfrac{\mathrm dy}{\mathrm dx} = x(3x-1)^2 \times (2(3x-1) + 9x) \\\\ \boxed{\dfrac{\mathrm dy}{\mathrm dx} = x(3x-1)^2(15x-2)}

You could also use logarithmic differentiation, which involves taking logarithms of both sides and differentiating with the chain rule.

On the right side, the logarithm of a product can be expanded as a sum of logarithms. Then use other properties of logarithms to simplify

\ln(y) = \ln\left(x^2(3x-1)^3\right) \\\\ \ln(y) =  \ln\left(x^2\right) + \ln\left((3x-1)^3\right) \\\\ \ln(y) = 2\ln(x) + 3\ln(3x-1)

Differentiate both sides and you end up with the same derivative:

\dfrac1y\dfrac{\mathrm dy}{\mathrm dx} = \dfrac2x + \dfrac9{3x-1} \\\\ \dfrac1y\dfrac{\mathrm dy}{\mathrm dx} = \dfrac{15x-2}{x(3x-1)} \\\\ \dfrac{\mathrm dy}{\mathrm dx} = \dfrac{15x-2}{x(3x-1)} \times x^2(3x-1)^3 \\\\ \dfrac{\mathrm dy}{\mathrm dx} = x(15x-2)(3x-1)^2

7 0
2 years ago
Please help me if possible ​
qaws [65]

Answer:

obtuse triangle

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
What the answer pls…
Valentin [98]
The slope is a positive and constant increase of 1 over 2 and the y - intercept is located at 8 due to the existing amount of snow on the ground. hope this helped you might want to word it a little differently.
3 0
3 years ago
"In the 45°-45°-90° triangle above, r = 10. Find the other two lengths.
Basile [38]

We are given a right triangle that has angles of 45°-45°-90°. This would indicate that the triangle is an isosceles type. We can use some trigonemetric functions to solve for the other legs. We do as follows:

 

sin 45 = p / 10

p = 5√2

 

cos 45 = q /10

q = 5√2

 

<span>Hope this answers the question. Have a nice day.</span>

5 0
3 years ago
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