All categories

3 years ago

Help me PLEASEEEEEEEEEEEEE

Mathematics

1 answer:

steposvetlana [31]3 years ago

4 0

Answer: d

Step-by-step explanation:

its d

You might be interested in

Picture above ^^^^^^^

Nitella [24]

Answer:

Step-by-step explanation:

Since this is a test/hw, I'll give a hint.

This problem at first can seem a bit difficult with q's and power's everywhere.

But let's take a step backward. A power is when your mutiplying something by itself again and again.

Ex: 3^3=3 times 3 times 3

But what if we had something liiiike this:

(3^3)2

In this case its now

(3 times 3 times 3)^2, so its "techinicaly" (27)^2. And you would a fairly large number, which I'm to lazy to solve. But that's not the point.

We've seen what a power is deconstructed, and what a power is. Because my explantion probably confused you more than it helped, I'll give an example.

(2^2)^2=(2 times 2)^2=(4^2=16=2^4

However, there is a shorter way to solve it.

(2^2)^2=2^(2 times 2)=2^4

Hope this helps.

8 0

3 years ago

How are these triangles congruent??????

Soloha48 [4]

Answer:

SAS Test

Step-by-step explanation:

6 0

3 years ago

The green triangle is a dilation of the red triangle with a scale factor of s=1/3 and the center of dilation is at the point (4,

klasskru [66]

Given:

The scale factor is $s=\dfrac{1}{3}$ and the center of dilation is at the point (4,2).

Red is original figure and green is dilated figure.

To find:

The coordinates of point C' and point A.

Solution:

Rule of dilation: If a figure is dilated with a scale factor k and the center of dilation is at the point (a,b), then

$(x,y)\to (k(x-a)+a,k(y-b)+b)$

According to the given information, the scale factor is $\dfrac{1}{3}$ and the center of dilation is at (4,2).

$(x,y)\to (\dfrac{1}{3}(x-4)+4,\dfrac{1}{3}(y-2)+2)$ ...(i)

Let us assume the vertices of red triangle are A(m,n), B(10,14) and C(-2,11).

Using (i), we get

$C(-2,11)\to C'(\dfrac{1}{3}(-2-4)+4,\dfrac{1}{3}(11-2)+2)$

$C(-2,11)\to C'(\dfrac{1}{3}(-6)+4,\dfrac{1}{3}(9)+2)$

$C(-2,11)\to C'(-2+4,3+2)$

$C(-2,11)\to C'(2,5)$

Therefore, the coordinates of Point C' are C'(2,5).

We assumed that point A is A(m,n).

Using (i), we get

$A(m,n)\to A'(\dfrac{1}{3}(m-4)+4,\dfrac{1}{3}(n-2)+2)$

From the given figure it is clear that the image of point A is (8,4).

$A'(\dfrac{1}{3}(m-4)+4,\dfrac{1}{3}(n-2)+2)=A'(8,4)$

On comparing both sides, we get

$\dfrac{1}{3}(m-4)+4=8$

$\dfrac{1}{3}(m-4)=8-4$

$(m-4)=3(4)$

$m=12+4$

$m=16$

And,

$\dfrac{1}{3}(n-2)+2=4$

$\dfrac{1}{3}(n-2)=4-2$

$(n-2)=3(2)$

$n=6+2$

$n=8$

Therefore, the coordinates of point A are (16,8).

5 0

3 years ago

I need help please and thank you

grin007 [14]

Answer:30

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Jimmy’s new cell phone cost him $49.99 when he signed a 2 year plan, which was 75% off the original place. What was the original

Sati [7]

If Jimmy's new cell cost him $49.99 with a 75% discount, then $49.99 is 100%−75%=25% of the original price.

The original price of the cell phone was $199.96.

6 0

4 years ago