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3 years ago

Question in picture. Thanks!

Mathematics

1 answer:

KonstantinChe [14]3 years ago

8 0

2x^2+4x+2
Carry the 2 over to the 3 behind the 4 and multiply so 2times2 is 4. So i new equation is.
X^2+4x+4
Now you have to find the factors of 4 which is 1,4 and 2,2
So now you have to see which one adds up to 4 so 1+4 is 5 so cross that out and 2+2=4.
So the two number is 2 and 2 now you have to bring down the x^2 and the signs.
(X+2) (X+2) and thats your answer

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A packing box is being filled with packing peanuts. The box measures 1 1/2 feet by 1 1/2 feet by 2 feet. How many cubic feet of

Stolb23 [73]

Answer: $4.5\ ft^3$

Step-by-step explanation:

Given

Measurement of box $1\frac{1}{2}\times 1\frac{1}{2}\times 2$

suppose Length $L=1\frac{1}{2}=\frac{3}{2}\ ft$

Breadth $B=1\frac{1}{2}=\frac{3}{2}\ ft$

height $H=2\ ft$

Volume of Cuboid $=L\times B\times H$

Volume of Packaging box $V=\frac{3}{2}\times \frac{3}{2}\times 2$

$V=\frac{9}{2}\ ft^3$

4 0

4 years ago

a triangle has an area of 54 m2 and a height of 9m how long is the base of the triangle entered your answer in the box

Delvig [45]

Answer:
base of the triangle = 12 m

Explanation:
Area of the triangle = 1/2 * base * height
We are given that:
area = 54 m^2
height = 9 m

Substitute in the equation to get the base as follows:
Area of the triangle = 1/2 * base * height
54 = 1/2 * base * 9
108 = 9*base
base = 12 m

Hope this helps :)

7 0

3 years ago

Read 2 more answers

1. (5pts) Find the derivatives of the function using the definition of derivative.

andreyandreev [35.5K]

2.8.1

$f(x) = \dfrac4{\sqrt{3-x}}$

By definition of the derivative,

$f'(x) = \displaystyle \lim_{h\to0} \frac{f(x+h)-f(x)}{h}$

We have

$f(x+h) = \dfrac4{\sqrt{3-(x+h)}}$

and

$f(x+h)-f(x) = \dfrac4{\sqrt{3-(x+h)}} - \dfrac4{\sqrt{3-x}}$

Combine these fractions into one with a common denominator:

$f(x+h)-f(x) = \dfrac{4\sqrt{3-x} - 4\sqrt{3-(x+h)}}{\sqrt{3-x}\sqrt{3-(x+h)}}$

Rationalize the numerator by multiplying uniformly by the conjugate of the numerator, and simplify the result:

$f(x+h) - f(x) = \dfrac{\left(4\sqrt{3-x} - 4\sqrt{3-(x+h)}\right)\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{\left(4\sqrt{3-x}\right)^2 - \left(4\sqrt{3-(x+h)}\right)^2}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{16(3-x) - 16(3-(x+h))}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{16h}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)}$

Now divide this by <em>h</em> and take the limit as <em>h</em> approaches 0 :

$\dfrac{f(x+h)-f(x)}h = \dfrac{16}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ \displaystyle \lim_{h\to0}\frac{f(x+h)-f(x)}h = \dfrac{16}{\sqrt{3-x}\sqrt{3-x}\left(4\sqrt{3-x} + 4\sqrt{3-x}\right)} \\\\ \implies f'(x) = \dfrac{16}{4\left(\sqrt{3-x}\right)^3} = \boxed{\dfrac4{(3-x)^{3/2}}}$

3.1.1.

$f(x) = 4x^5 - \dfrac1{4x^2} + \sqrt[3]{x} - \pi^2 + 10e^3$

Differentiate one term at a time:

• power rule

$\left(4x^5\right)' = 4\left(x^5\right)' = 4\cdot5x^4 = 20x^4$

$\left(\dfrac1{4x^2}\right)' = \dfrac14\left(x^{-2}\right)' = \dfrac14\cdot-2x^{-3} = -\dfrac1{2x^3}$

$\left(\sqrt[3]{x}\right)' = \left(x^{1/3}\right)' = \dfrac13 x^{-2/3} = \dfrac1{3x^{2/3}}$

The last two terms are constant, so their derivatives are both zero.

So you end up with

$f'(x) = \boxed{20x^4 + \dfrac1{2x^3} + \dfrac1{3x^{2/3}}}$

8 0

2 years ago

What is the distance between points M and N?

Free_Kalibri [48]

By using <em>triangle</em> properties and the law of the cosine twice, we find that the distance between points M and N is approximately 9.8 meters.

<h3>How to determine the distance between two points</h3>

In this problem we must determine the distance between two points that are part of a triangle and we can take advantage of properties of triangles to find it. First, we determine the measure of angle L by the law of the cosine:

$\cos L = \frac{(19.6\,m)^{2}-(14.8\,m)^{2}-(21.4\,m)^{2}}{-2\cdot (14.8\,m)\cdot (21.4\,m)}$

L ≈ 62.464°

Then, we get the distance between points M and N by the law of the cosine once again:

$MN = \sqrt{(7.4\,m)^{2}+(10.7\,m)^{2}-2\cdot (7.4\,m)\cdot (10.7\,m)\cdot \cos 62.464^{\circ}}$

MN ≈ 9.8 m

By using <em>triangle</em> properties and the law of the cosine twice, we find that the distance between points M and N is approximately 9.8 meters.

To learn more on triangles: brainly.com/question/2773823

#SPJ1

6 0

2 years ago

Use decimals to give the ordered pair for point U. Remember to putyour odered pairs in parenthesis.

Gnoma [55]

Answer:

(0,2.5)

Step-by-step explanation:

3 0

3 years ago