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3 years ago

Please help with at least one of these( I accept 1 answer if you have it; make sure you are 100% positive its correct though!)

Mathematics

1 answer:

Tema [17]3 years ago

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Tim is 5 years older than Melissa. The sum of their ages is 21. This system is represented by the equations: t = 5 + m t + m = 2

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3 years ago

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2 years ago

Which equation represents a proportional relationship? Oy=2(x + 1) Oy=2 o y = 4x + 3 o y = 32x y

kherson [118]

Answer:

y=32x shows a proportional relationship.

Step-by-step explanation:

We need to find an equation that shows the proportional relationship.

If x and y are proportional to each other, it will look like,

y = mx

Where

m is constant of proportionality

Out of 4 option, option (4) is y=32x, it shows the proportional relationship where the value of m is equal to 32.

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3 years ago

Simplify the expression below. (5x)2

Anarel [89]

10x ..................

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3 years ago

Please help !!! I’ll mark brainliest

frozen [14]

A) State the chain rule for integration

Ans. The chain rule for integration is also known as " Integration by substitution "

Integration by substitution is taken in order to make integration solve easily in few steps.

For, $I = \int\limits (x+2)^{2} \,dx$

Instead of expanding term $(x+2)^{2}$

With substitution of $u = (x+2)$ and $du= 1 dx$

We simplified the integration as

$I = \int\limits (u)^{2} \, du$

$I = \frac{(u)^{3}}{3}+C$

By replacaing value of u=x+2

$I = \frac{(x+2)^{3}}{3}+C$

B) State the rule of differentiation for the sine function.

Ans. We know that $\frac{d}{dx}Sinx dx = Cosx$

C) Find the indefinite integral using substitution.

Ans.

Given, $I = \int\limits {\frac{Cos14x}{Sin14x} } \, dx$

Take y = Sin14x

Differentiating both side

$dy=14Cos14x dx$

$\frac{dy}{14} = Cos14x\, dx$

Substituting values in integration,

$I = \int\limits {\frac{Cos14x}{Sin14x} } \, dx$

$I = \int\limits {\frac{1}{y} } \,\frac{dy}{14}$

$I = \frac{1}{14}\int\limits {\frac{1}{y} } \,dy$

$I = \frac{1}{14} lny + C$

Replacing values in the integration

$I = \frac{1}{14} ln(14Sin14x) + C$

D)Check your work by taking a derivative of your answer from part C.

Ans.

Answer for Part C is $I = \frac{1}{14} ln(14Sin14x) + C$

Differentiating the answer

we get,

$=\frac{1}{14}\frac{d}{dx}[ ln(Sin14x) + C]\\=\frac{1}{14}\frac{1}{Sin14x} \frac{d}{dx}(Sin14x)+ \frac{d}{dx}C\\=\frac{1}{14}\frac{1}{Cos14x}(14Cos14x)\\=\frac{Cos14x}{Sin14x} \\ =I$

6 0

3 years ago