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3 years ago

Please help :(( I’m having a hard time with these

Mathematics

1 answer:

pantera1 [17]3 years ago

3 0

Answer:

Step-by-step explanation:

what you do on one side of the equal sign, you do to the other

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If the radius is 18 cm and the angle at centre of circle is 120° then the

PolarNik [594]

The area of a sector is 339.336 cm²

The area of a sector is defined to be as the region of space bounded by a boundary from the center of the circle.

It can be determined by using the formula:

$\mathbf{A = \dfrac{\theta }{360}\times \pi r^2 }$

Given that;

Radius r = 18 cm
Angle θ = 120°

Then. the area of the sector can be calculated as:

$\mathbf{A = \dfrac{120}{360}\times \pi (18)^2 }$

$\mathbf{A = \dfrac{1}{3}\times 3.142 \times 324 }$

A = 339.336 cm²

Learn more about the area of a sector here:

brainly.com/question/7512468?referrer=searchResults

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2 years ago

Answer the question in the picture. If its right I’ll mark brainliest

Kisachek [45]

Answer:

The answer is c!

Step-by-step explanation:

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3 years ago

Find the slope of the line that passes through (6, 7) and (3, 9).

Karolina [17]

Answer:

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Step-by-step explanation:

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3 years ago

Venus is about 7.2 x 10-1 AU from the Sun. Mars is about 1.5 AU from the Sun.

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3 years ago

<img src="https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdx%7D%20%5Cint%20t%5E2%2B1%20%5C%20dt" id="TexFormula1" title="\frac{d}{dx} \

Kisachek [45]

Answer:

$\displaystyle{\frac{d}{dx} \int \limits_{2x}^{x^2} t^2+1 \ \text{dt} \ = \ 2x^5-8x^2+2x-2$

Step-by-step explanation:

$\displaystyle{\frac{d}{dx} \int \limits_{2x}^{x^2} t^2+1 \ \text{dt} = \ ?$

We can use Part I of the Fundamental Theorem of Calculus:

$\displaystyle\frac{d}{dx} \int\limits^x_a \text{f(t) dt = f(x)}$

Since we have two functions as the limits of integration, we can use one of the properties of integrals; the additivity rule.

The Additivity Rule for Integrals states that:

$\displaystyle\int\limits^b_a \text{f(t) dt} + \int\limits^c_b \text{f(t) dt} = \int\limits^c_a \text{f(t) dt}$

We can use this backward and break the integral into two parts. We can use any number for "b", but I will use 0 since it tends to make calculations simpler.

$\displaystyle \frac{d}{dx} \int\limits^0_{2x} t^2+1 \text{ dt} \ + \ \frac{d}{dx} \int\limits^{x^2}_0 t^2+1 \text{ dt}$

We want the variable to be the top limit of integration, so we can use the Order of Integration Rule to rewrite this.

The Order of Integration Rule states that:

$\displaystyle\int\limits^b_a \text{f(t) dt}\ = -\int\limits^a_b \text{f(t) dt}$

We can use this rule to our advantage by flipping the limits of integration on the first integral and adding a negative sign.

$\displaystyle \frac{d}{dx} -\int\limits^{2x}_{0} t^2+1 \text{ dt} \ + \ \frac{d}{dx} \int\limits^{x^2}_0 t^2+1 \text{ dt}$

Now we can take the derivative of the integrals by using the Fundamental Theorem of Calculus.

When taking the derivative of an integral, we can follow this notation:

$\displaystyle \frac{d}{dx} \int\limits^u_a \text{f(t) dt} = \text{f(u)} \cdot \frac{d}{dx} [u]$
where u represents any function other than a variable

For the first term, replace $\text{t}$ with $2x$ , and apply the chain rule to the function. Do the same for the second term; replace

$\displaystyle-[(2x)^2+1] \cdot (2) \ + \ [(x^2)^2 + 1] \cdot (2x)$

Simplify the expression by distributing $2$ and $2x$ inside their respective parentheses.

$[-(8x^2 +2)] + (2x^5 + 2x)$
$-8x^2 -2 + 2x^5 + 2x$

Rearrange the terms to be in order from the highest degree to the lowest degree.

$\displaystyle2x^5-8x^2+2x-2$

This is the derivative of the given integral, and thus the solution to the problem.

6 0

3 years ago