"The sum of the abscissa and the ordinate is six"?

HELPPP FIND THE AREA !!

Rashid [163]

Answer:

To find the area, you have to multiply its height by its width.

Step-by-step explanation:

7 0

3 years ago

What is the product of 3x(x2 4)? x2 3x 4 3x3 4 3x3 12x 3x2 12x.

iVinArrow [24]

Answer:

3x(x2 + 4)

Then use the distributive property

SO the final answer would be: 3x^3 +12x

Step-by-step explanation:

5 0

2 years ago

Ted is making trail mix for a party. He mixes 1 1/2 cups of nuts, 1/4 cup of raisins, and 1/4 cup of pretzels. How many cups of

cupoosta [38]

Answer:

Ted will need $3\frac{3}{4}$ cups of pretzels to make 15 cups of trail mix.

Step-by-step explanation:

Ted is making trail mix for a party. He mixes 1 1/2 cups of nuts, 1/4 cup of raisins, and 1/4 cup of pretzels.

So, 1/4 cup of pretzels to make 1 trail mix

x cups of pretzels to make 15 trail mix

Using the ratio and proportional

∴ x = (1/4) * 15 = 3.75 = $3\frac{3}{4}$ cups.

<u>Ted will need </u> $3\frac{3}{4}$ <u> cups of pretzels to make 15 cups of trail mix.</u>

4 0

3 years ago

What is the total cost of an item priced $4.95 if there is 5% sales tax?

Paha777 [63]

4.95+4.95×0.05=5.20
5 dollars 20 cents

7 0

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

2 years ago