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Sergio [31]
3 years ago
15

The point (7, 0) is reflected over the x-axis. What do you notice about the coordinates of the reflected point? Why is this?

Mathematics
1 answer:
goblinko [34]3 years ago
5 0

Answer:

The coordinate stayed the same

Step-by-step explanation:

because when you reflect on the x-axis, the x-axis stays the same while the y-axis gets flipped. in this case, the coordinate is on the y-axis, or the y-axis is 0, and you can't flip 0 to get an opposite. the opposite of 0 is 0, so he coordinate stayed the same.

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A closed cylindrical tank of radius 1.50m and height 300cm, is made from a metal sheet. How much sheet is required?
Anika [276]

Answer: 13.5π≈42.41

Step-by-step explanation:

Basically it is asking for the circumference of this cylinder.

#1 Find the circumference of the circle base

Formula to use: 2πr

2πr=2π(1.5)=3π

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

#2 Find the area of the circle base

Formula to use: πr²

πr²=π(1.5)²=2.25π

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

#3 Find the area of body of the cylinder

Formula to use: lw

300cm=3m

l (length)= 3π (since the body of the cylinder the basing on the circumference of the circle base)

w (width)= 3

lw=3π×3m=9π

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

#4 Add all of them together

There are 2 base and one body

2.25π×2+9π=13.5π

4 0
3 years ago
What are the solutions of the equation 6x2 + x-1=0 ?
Aleksandr [31]

Answer:

<u>Options B and D</u>

Step-by-step explanation:

6x² + x - 1 = 0

6x² + 3x - 2x - 1 = 0

3x(2x + 1) - 1(2x + 1) = 0

(3x - 1)(2x + 1)

x = 1/3 and -1/2

<u>Options B and D</u>

6 0
2 years ago
How to find imaginary zeros anf real zeros of F(x)=-4x^5-8x^3+12x​
Fofino [41]

Answer:

x=\{0, -1, 1, -i\sqrt{3}, i\sqrt{3}\}

Step-by-step explanation:

We are given the function:

f(x)=-4x^5-8x^3+12x

And we want to finds its zeros.

Therefore:

0=-4x^5-8x^3+12x

Firstly, we can divide everything by -4:

0=x^5+2x^3-3x

Factor out an x:

0=x(x^4+2x^2-3)

This is in quadratic form. For simplicity, we can let:

u=x^2

Then by substitution:

0=x(u^2+2u-3)

Factor:

0=x(u+3)(u-1)

Substitute back:

0=x(x^2+3)(x^2-1)

By the Zero Product Property:

x=0\text{ and } x^2+3=0\text{ and } x^2-1=0

Solving for each case:

x=0\text{ and } x=\pm\sqrt{-3}\text{ and } x=\pm\sqrt{1}

Therefore, our real and complex zeros are:

x=\{0, -1, 1, -i\sqrt{3}, i\sqrt{3}\}

4 0
3 years ago
FINALS HELP SORRY!!!!!!
monitta
I believe the 4th one
an=145+an-1 and a1=20
4 0
3 years ago
Read 2 more answers
The radius of the circular track was 157.64 feet find the circumference of the track
DerKrebs [107]
Before we solve for anything, let's remember what our formula is.

The circumference of a circle is 2PiR (2 x Pi x R).

Our radius is 157.64.
Pi is 3.14 (estimated).

Let's plug our numbers into the formula.

Circumference = 2 x 3.14 x 157.64.
2 x 3.14 = 6.28
6.28 x 157.64 = 989.979 (990 if estimated).

Your circumference is:
989.979 ft^2.

I hope this helps!
7 0
3 years ago
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