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3 years ago

PLEASE HELP !!!!!!!!!! WILL FOLLOW

Mathematics

1 answer:

antoniya [11.8K]3 years ago

8 0

Answer:

1/12

Step-by-step explanation:

2/3=8/12

2x4=8

3x4=12

1/4=3/12

1x3=3

4x3=12

8/12+3/12=11/12

1-11/12= 1/12

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During an Apollo moon landing, reflecting panels were placed on the moon. This allowed Earth-based astronomers to shoot laser be

rusak2 [61]

Answer:

The distance between the astronomers and the moon was $7.56*10^{8]$ meters.

Step-by-step explanation:

We have that the speed is the distance divided by the time, so:

$v = \frac{d}{t}$

In this problem, we have that:

The reflected laser beam was observed by the astronomers 2.52 s after the laser pulse was sent. This means that $t = 2.52$ .

If the speed of light is 3.00 times 10^8 m/s, what was the distance between the astronomers and the moon?

We have that $v = 3*10^{8}$ m/s.

We have to find d. So:

$v = \frac{d}{t}$

$d = vt$

$d = 3*10^{8}*2.52$

$7.56*10^{8]$

The distance between the astronomers and the moon was $7.56*10^{8]$ meters.

7 0

3 years ago

I need to know how to solve for the value of x

dimulka [17.4K]

Hey Katie

4^2x+3 = 1
We need to solve for x

log(4^2x+3) = log (1)
Now take log of
(2x +3) * log (4) = log (1)
2x + 3 = log(1)/log(4)
2x + 3 = 0
2x = 0 - 3
2x = -3
x = -3/2

I hope that's help !

8 0

2 years ago

Need help with number 6 plssssssssssassssa

Advocard [28]

C I think sorry I will try I will get other answers later

8 0

3 years ago

What are the solutions to the equation

frosja888 [35]

Answer:

$x_1=\frac{1}{4}+(\frac{\sqrt{7}}{4})i$ and $x_2=\frac{1}{4}-(\frac{\sqrt{7} }{4})i$

Step-by-step explanation:

You have the quadratic function $2x^2-x+1=0$ to find the solutions for this equation we are going to use Bhaskara's Formula.

For the quadratic functions $ax^2+bx+c=0$ with $a\neq 0$ the Bhaskara's Formula is:

$x_1=\frac{-b+\sqrt{b^2-4.a.c} }{2.a}$

$x_2=\frac{-b-\sqrt{b^2-4.a.c} }{2.a}$

It usually has two solutions.

Then we have $2x^2-x+1=0$ where a=2, b=-1 and c=1. Applying the formula:

$x_1=\frac{-b+\sqrt{b^2-4.a.c} }{2.a}\\\\x_1=\frac{-(-1)+\sqrt{(-1)^2-4.2.1} }{2.2}\\\\x_1=\frac{1+\sqrt{1-8} }{4}\\\\x_1=\frac{1+\sqrt{-7} }{4}\\\\x_1=\frac{1+\sqrt{(-1).7} }{4}\\x_1=\frac{1+\sqrt{-1}.\sqrt{7}}{4}$

Observation: $\sqrt{-1}=i$

$x_1=\frac{1+\sqrt{-1}.\sqrt{7}}{4}\\\\x_1=\frac{1+i.\sqrt{7}}{4}\\\\x_1=\frac{1}{4}+(\frac{\sqrt{7}}{4})i$

And,

$x_2=\frac{-b-\sqrt{b^2-4.a.c} }{2.a}\\\\x_2=\frac{-(-1)-\sqrt{(-1)^2-4.2.1} }{2.2}\\\\x_2=\frac{1-i.\sqrt{7} }{4}\\\\x_2=\frac{1}{4}-(\frac{\sqrt{7}}{4})i$

Then the correct answer is option C.

$x_1=\frac{1}{4}+(\frac{\sqrt{7}}{4})i$ and $x_2=\frac{1}{4}-(\frac{\sqrt{7} }{4})i$

3 0

2 years ago

Does — бр + 7 = 7 – бр have <br> No Solution, One Solution, or Infinitely

julsineya [31]

Answer:

Infinitely

Step-by-step explanation:

Hi,

Both sides of the equation are equal to each other, this means they're the exact same line and overlap. This is infinitely many solutions.

Hope this helps :)

4 0

2 years ago

Read 2 more answers