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Maslowich
3 years ago
8

Anyone Wanna help thanks

Mathematics
1 answer:
Oksi-84 [34.3K]3 years ago
6 0

Answer:

1. 8

2. 4

3a. -6 and -4

3b.  -1/2 and -5

Step-by-step explanation:

3a explanation:

Factor x^2+10x+24

(x+4)(x+6)

Set the factors equal to 0

x + 4 = 0

x + 6 = 0

-4 and -6

3b.

Factor left side of equation.

(2x+1)(x+5)=0

Set factors equal to 0.

2x + 1 = 0 or

x + 5 = 0

x =  −1/2

x = -5

-5 and -1/2

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If anybody answers these questions I will get them brainliest and if somebody else verifies that it is the correct answer I can
valina [46]

Part (a)

<h3>Answer: x^3 + 12x^2 + 47x + 60</h3>

--------------------------

Work Shown:

Let's say we have

  • L = length = x+5
  • W = width = x+4
  • H = height = x+3

To find the volume of the block, we multiply the length width and height.

So,

volume = (length)*(width)*(height)

V = L*W*H

V = LW*(x+3) .... replace H with (x+3)

V = LW*x + LW*3 .... distribute

V = Lx*(W) + 3L*(W)

V = Lx(x+4) + 3L(x+4) .... replace W with x+4

V = Lx^2 + 4Lx + 3Lx + 12L .... distribute

V = Lx^2 + 7Lx + 12L

V = x^2*(L) + 7x*(L) + 12*(L)

V = x^2*(x+5) + 7x*(x+5) + 12*(x+5) .... replace L with x+5

V = x^3 + 5x^2 + 7x^2 + 35x + 12x + 60 .... distribute

V = x^3 + 12x^2 + 47x + 60

x^3 + 12x^2 + 47x + 60 is the polynomial in standard form that represents the volume of the block.

=======================================================

Part (b)

<h3>Answer: 336 cubic feet</h3>

--------------------------

Work Shown:

We can plug x = 3 into the polynomial we found

V = x^3 + 12x^2 + 47x + 60

V = (3)^3 + 12(3)^2 + 47(3) + 60

V = 27 + 12(9) + 47(3) + 60

V = 27 + 108 + 141 + 60

V = 336

Or we could find the volume like this

V = L*W*H

V = (x+5)*(x+4)*(x+3)

V = (3+5)*(3+4)*(3+3)

V = (8)*(7)*(6)

V = 336

Either way, we get the same volume of 336 cubic feet.

3 0
3 years ago
Help me....................​
marta [7]

Answer:

2*2 = 4

4^4 = 16

16 > x + 4

Step-by-step explanation:

X is any real number less than 12

7 0
3 years ago
Find the angle measure to the nearest degree. <br> cos A = 0.7431<br> How do I do this?
svet-max [94.6K]

Answer:

Step-by-step explanation:

If you are looking for a missing angle measure, you use the 2nd button and the cos button.  Make sure, first off, that your calculator is in "degree" mode by hitting the "mode" button and making sure that the "degree" is highlighed and not the "radian".  Then hit "clear".  Once you know that you are in the correct mode, hit "2nd" then "cos" and you will see this on your screen:

cos^{-1}(

Inside the parenthesis you will enter your decimal, so it looks like this now:

cos^{-1}(.7431

You do NOT have to close the parenthesis, but you can if you want to.  Then hit "enter" to get that the angle that has a cosine of .7431 is 42.0038314 or, to the nearest degree, 42

6 0
3 years ago
Kenneth earns $450 for working 37 1/2 hours. If Kenneth works for 42 hours next week, how much will he earn. A.255.00. B.12.00.
iren2701 [21]

Answer:

Kenneth will earn about $ 504 the next week

Step-by-step explanation:

The first thing we will have to figure out is how much Kenneth earns per hour. Then we can get how much he will earn when he wors for 42 hours.

to get Kenneth's wage per hour, we will divide the sum he is paid in the first week by the number of hours he worked that week.

This is 450/ 37.5 = 12 dollars per hour.

Kenneth earns $12 per hour.

If Kenneth works for 42 hours, he will earn 42 X 12 = $ 504 the next week

5 0
3 years ago
Compute the surface area of the portion of the sphere with center the origin and radius 4 that lies inside the cylinder x^2+y^2=
Tom [10]

Answer:

16π

Step-by-step explanation:

Given that:

The sphere of the radius = x^2 + y^2 +z^2 = 4^2

z^2 = 16 -x^2 -y^2

z = \sqrt{16-x^2-y^2}

The partial derivatives of Z_x = \dfrac{-2x}{2 \sqrt{16-x^2 -y^2}}

Z_x = \dfrac{-x}{\sqrt{16-x^2 -y^2}}

Similarly;

Z_y = \dfrac{-y}{\sqrt{16-x^2 -y^2}}

∴

dS = \sqrt{1 + Z_x^2 +Z_y^2} \ \ . dA

=\sqrt{1 + \dfrac{x^2}{16-x^2-y^2} + \dfrac{y^2}{16-x^2-y^2} }\ \ .dA

=\sqrt{ \dfrac{16}{16-x^2-y^2}  }\ \ .dA

=\dfrac{4}{\sqrt{ 16-x^2-y^2}  }\ \ .dA

Now; the region R = x² + y² = 12

Let;

x = rcosθ = x; x varies from 0 to 2π

y = rsinθ = y; y varies from 0 to \sqrt{12}

dA = rdrdθ

∴

The surface area S = \int \limits _R \int \ dS

=  \int \limits _R\int  \ \dfrac{4}{\sqrt{ 16-x^2 -y^2} } \ dA

= \int \limits^{2 \pi}_{0} } \int^{\sqrt{12}}_{0} \dfrac{4}{\sqrt{16-r^2}} \  \ rdrd \theta

= 2 \pi \int^{\sqrt{12}}_{0} \ \dfrac{4r}{\sqrt{16-r^2}}\ dr

= 2 \pi \times 4 \Bigg [ \dfrac{\sqrt{16-r^2}}{\dfrac{1}{2}(-2)} \Bigg]^{\sqrt{12}}_{0}

= 8\pi ( - \sqrt{4} + \sqrt{16})

= 8π ( -2 + 4)

= 8π(2)

= 16π

4 0
3 years ago
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