All categories

3 years ago

The triangular prisms shown below are similor. Whet are the values of x and y?

1 answer:

OverLord2011 [107]3 years ago

5 0

Answer:To find the volume of a rectangular prism, multiply its 3 dimensions: length x width x height. The volume is expressed in cubic units.

Step-by-step explanation:

You might be interested in

Tanikwa started a new exercise program using the indoor track. In her first workout she ran 8 laps for every 4 laps that she wal

Elina [12.6K]

Answer:

The correct answer would be 2:1

Step-by-step explanation:

In order to find this, start with the ratio as it is written and sub in the values. Then simplify to the lowest possible terms.

Ran:Walker

8:4

2:1

6 0

3 years ago

Which is the completely factored form of 3xsquared<br> -12x-15?

tekilochka [14]

Answer:

3(x - 5)(x + 1).

Step-by-step explanation:

3x^2 - 12x - 15

Dividing through by 3:

= 3(x^2 - 4x - 5)

We need 2 numbers whose sum is -4 and whose product = -5. That is -5 and +1 , so the factors are:

3(x - 5)(x + 1).

7 0

3 years ago

I need help finding the perimeter

Romashka [77]

Measure the sides. And when you get all of the sides add them I'll and get the perimeter I think.

3 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

If DM = 35, what is the value of r?

Gennadij [26K]

Answer:

$r = 11$

Step-by-step explanation:

Given

$DM = 35$

$DG = r + 5$

$GM = 3r - 14$

Required

Find r

From the question, we understand that G is a point between D and M:

This implies that:

$DM = DG + GM$

Substitute values for DM, DG and GM

$35 = r + 5 + 3r - 14$

Collect Like Terms

$r + 3r = 35 - 5 + 14$

$4r = 44$

Solve for r

$r = 44/4$

$r = 11$

3 0

3 years ago