A triangle has an area of 90 and a height of 10. What is the base

HELPP PLEASEEE!!<br> (look at the photo)

Angelina_Jolie [31]

Answer:

B

Step-by-step explanation:

If you notice, since concert tickets is 1, the cost is $23. So we are finding an option that satifies that. The only option is B.

Hope you get a 100. Please hit the crown if correct :D

5 0

2 years ago

Read 2 more answers

Find the volume of each cylinder. Use 3.14 for pi. Round your answer to the nearest tenth

otez555 [7]

Answer:

V ≈471.24 mm^3

Step-by-step explanation:

The formula for cylinder volume is πr^2 x h, so ((π x 25) x h). That's just 25π x 6. That is about 471.238898, which rounded is almost 471.24. Or, in terms of π, you could leave your answer as 150π mm^3

8 0

3 years ago

Issac put $5 in a shoe box every day for the months of april, may, and june. He then spent 75% of the money on baseball cards. H

igor_vitrenko [27]

As of 2021:
April-$150
May-$155
June-$150
Total-$455
Spent 75%=455/100x75
=341.25
455-341.25=$113.75
Round off to 3sf: $114
Have a good day

3 0

2 years ago

Here is a uniform probability distribution.

Darya [45]

a) $P(4\le X\le6)=\dfrac25$ because it is equal to the area of the shaded region between X=4 and X=6, and the probability that X falls within some interval is given by the area under the PDF.

b) $a=3$ because the shaded region is a rectangle of height 1/5 (by virtue of X following a uniform distribution over the interval [2, 7], which has length 5).

4 0

2 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