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Lisa [10]
2 years ago
10

What can 16 and 4 both be divided by​

Mathematics
1 answer:
ryzh [129]2 years ago
3 0

Answer:

4

Step-by-step explanation:

they can both be divided by 4

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2/5% as a fraction in it's simplest form​
Llana [10]

Answer:

Here is the ans...hope it helps:)

4 0
2 years ago
rish is sorting bolts for her home repair business. She has an 8-pound bag of bolts. If each kitchen her team repairs requires 1
Reptile [31]
I hope you mean 1/5 when you say 1 5. If that's what you mean, then both, Joshua and Melanie, are correct because, both of their equations are the same. 

Joshua's - <span>8 ÷ 1/5 = 40 (Since we can divide a number by fraction, we use this rule: keep, change flip. When we use that Joshua's equation turns into 8 * 5/1 = 8*5 = 40, same as Melanie's.)</span>
5 0
3 years ago
Which is the equation of a circle with its center at (-4,2) and a radius of 4?
Snowcat [4.5K]

Answer:

(x +4)^2 + (y-2)^2 = 16

Step-by-step explanation:

We can write the equation for a circle with the formula

(x-h)^2 + (y-k)^2 = r^2

Where (h,k) is the center and r is the radius

We know the center is at (-4,2) and the radius is 4

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

(x +4)^2 + (y-2)^2 = 16

4 0
3 years ago
GIVING BRAINILEST-
Nutka1998 [239]

Answer:

To construct a dilation, we need the center of dilation, which is the point from which we make the image smaller or larger. We also need a scale factor, which is the number that determines how large or small we are making the new image. If the scale factor is greater than 1, we have an enlargement, because the new image will be larger than the original image. If the scale factor is less than 1, we have a reduction because the new image will be smaller than the original image.

Enlargement

Step-1

Let's dilate triangle PQR by a scale factor of 2 with a center of dilation at point C.

Step-2

First, we will use a straightedge to connect the center of dilation C to each vertex. We are going to extend those lines across the whole page.

Step-3

Next, we'll place the point of the compass on the center of dilation C and the pencil on vertex Q and measure that distance.

Step-4

Now, without changing the size of the compass, we will move the point of the compass to vertex Q, and make a mark on the line that is extended through Q.

Step-5

Notice that since we measured the distance from C to Q and then used that same distance to mark the line, we have now created a new point that is twice the distance from C that Q was.

We now repeat the process for each of the other vertices of the triangle.

Step-6

These marks represent the vertices of our dilated image. We often use the same letters with an apostrophe to show that it is the corresponding vertex.

Step-7

Finally, we use the straightedge to connect all of the new vertices.

Now,See how triangle P

′

Q

′

R

′

is twice the size of the original triangle.

Reduction

Step-1

Given triangle PQR, now let's dilate the image with a center of dilation at C and with a scale factor of

2

1

Step-2

Using the straightedge, we will connect all of the vertices to the center of dilation C. In this case, since our scale factor is less than 1, we are reducing the size of the triangle.

Step-3

The green lines are the new lines that we have drawn.

Since we have to reduce the size of the triangle by half, we want to cut the distance in half from each vertex to the center of dilation. In order to do this, we will set our compass to a distance that is more than halfway between vertex P and point C but less than the total distance.

Step-4

Keeping the compass at that same distance, start with the point of the compass on vertex P, then draw a curve that intersects the line between P and C. Next, keeping the compass at the same distance, place the point of the compass on C and draw a curve that intersects the line between P and C and intersects the previously drawn curve.

Step-5

Now, we will take the straightedge and connect the points where the curves overlap.

4 0
3 years ago
Frank cut a 52 inch string into 3 pieces. The first piece is 20 inches long. The second piece is at least 12 inches long, but no
Eddi Din [679]
We need to figure out how much string would be left, after taking away the first two pieces.
We know that the first piece is 20 inches long, so we can say that there is 52-20 inches left, or 32 inches.

The second piece is between 12 and 18 inches, meaning that there would be between 32-12 and 32-18 inches left for the third piece, or 20 and 14 inches. This means that the third piece would be at least 14 inches long, but no more than 20, since we don’t have more string than that (20+12+20=52, and 20+14+18=52)

So we can say that x is greater or equal to 14, but less than or equal to 20, or:

14<=x<=20 (“<=“ is written like a normal “<“ sign with a line _ right under it)
4 0
3 years ago
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