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3 years ago

Answer above question in image

Mathematics

1 answer:

slava [35]3 years ago

4 0

Answer:

Option B you answer

Step-by-step explanation:

Counter clockwise is you going --->

so if the shape is on the left side of the D

then 90 degrees put it Straight up and down

Having point B sticking out

Ps Option C not it cause that is 180 degree rotation same as A but its clockwise

Also D is clock wise not COUNTER CLOCKWISE

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Solve for x. Round your answer to the nearest tenth if necessary.

eduard

Answer: 15.9

Step-by-step explanation:

7 0

2 years ago

Help!! 50 points and brainliest!

Viktor [21]

Answer:

Second choice:

$x=2t$

$y=4t^2+4t-3$

Fifth choice:

$x=t+1$

$y=t^2+4t$

Step-by-step explanation:

Let's look at choice 1.

$x=t+1$

$y=t^2+2t$

I'm going to subtract 1 on both sides for the first equation giving me $x-1=t$ . I will replace the $t$ in the second equation with this substitution from equation 1.

$y=(x-1)^2+2(x-1)$

Expand using the distributive property and the identity $(u+v)^2=u^2+2uv+v^2$ :

$y=(x^2-2x+1)+(2x-2)$

$y=x^2+(-2x+2x)+(1-2)$

$y=x^2+0+-1$

$y=x^2$

So this not the desired result.

Let's look at choice 2.

$x=2t$

$y=4t^2+4t-3$

Solve the first equation for $t$ by dividing both sides by 2:

$t=\frac{x}{2}$ .

Let's plug this into equation 2:

$y=4(\frac{x}{2})^2+4(\frac{x}{2})-3$

$y=4(\frac{x^2}{4})+2x-3$

$y=x^2+2x-3$

This is the desired result.

Choice 3:

$x=t-3$

$y=t^2+2t$

Solve the first equation for $t$ by adding 3 on both sides:

$x+3=t$ .

Plug into second equation:

$y=(x+3)^2+2(x+3)$

Expanding using the distributive property and the earlier identity mentioned to expand the binomial square:

$y=(x^2+6x+9)+(2x+6)$

$y=(x^2)+(6x+2x)+(9+6)$

$y=x^2+8x+15$

Not the desired result.

Choice 4:

$x=t^2$

$y=2t-3$

I'm going to solve the bottom equation for $t$ since I don't want to deal with square roots.

Add 3 on both sides:

$y+3=2t$

Divide both sides by 2:

$\frac{y+3}{2}=t$

Plug into equation 1:

$x=(\frac{y+3}{2})^2$

This is not the desired result because the $y$ variable will be squared now instead of the $x$ variable.

Choice 5:

$x=t+1$

$y=t^2+4t$

Solve the first equation for $t$ by subtracting 1 on both sides:

$x-1=t$ .

Plug into equation 2:

$y=(x-1)^2+4(x-1)$

Distribute and use the binomial square identity used earlier:

$y=(x^2-2x+1)+(4x-4)$

$y=(x^2)+(-2x+4x)+(1-4)$

$y=x^2+2x+-3$

$y=x^2+2x-3$ .

This is the desired result.

3 0

4 years ago

Read 2 more answers

New probabilities that have been found using Bayes' theo- rem are called a. prior probabilities. b. posterior probabilities. c.

Sphinxa [80]

Answer:

b). posterior probabilities.

Step-by-step explanation:

Posterior Probabilities are described as the updated probabilities which are estimated employing the theorem proposed by Baye. According to his theorem, an association is developed between probability and the newly received information. <u>When the previously existing probabilities are amalgamated with this updated information, it gives the existing probabilities' value a correction, update, or revision namely 'posterior probabilities</u>.' Thus, it is the probability that asserts the truth of a hypothesis and hence, <u>option b</u> is the correct answer.

6 0

3 years ago

Find the slope of (1,1) and (3,1)

Dafna11 [192]

The slope is 0/2 (0) because they are on the same x line.

8 0

3 years ago

I need help please the friends would greatly appreciate it. plotted in the complex plane ... :)

zlopas [31]

Attached is a picture showing the plot in the plane.
Note that for a complex number , x + iy.
The real part of the number is put on the x axis and the imaginary part is put on the y axis.

So plotting:
A = 4 + 1i
B = 9 + 7i
C = 5 + 8i

See picture.

7 0

3 years ago

Answer above question in image​

Answer above question in image