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Sidana [21]
3 years ago
11

10^4 divided by 10^19 =?

Mathematics
1 answer:
Len [333]3 years ago
6 0

Answer:

0000000000.00001

Step-by-step explanation:

7ycf6gffty

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Solve the following system of equations graphically on the set of axes below. y=3x−3 , x+y=5
topjm [15]

Answer:

x=2, y=3

Step-by-step explanation:

First, graph both equations on a rectangular coordinate system (see attached screenshot). Then, simply figure out where these two lines intersect. Since they intersect at (2,3), x=2 and y=3.

Plugging these values in the equations for x and y, you can see if these are the correct answers:

(3)=3(2)-3, (2)+(3)=5

Since both of these are true, you have the right answer.

Hope this helps!

6 0
3 years ago
Kara mixes different colors of paint to create new colors. The table shows the amount of paint Kara mixes per batch.
Sergio [31]

Answer:

D. Batch 5.

Step-by-step explanation:

The batch should have the same proportion of blue to white to yellow.

In Batch 1, there are two parts of blue, 1.5 parts of white, and 1 part of yellow.

In Batch 5, there are four parts of blue, 3 parts of white, and 2 parts of yellow.

4 / 2 = 2

3 / 2 = 1.5

2 / 2 = 1

Since the proportions are equal to those found in Batch 1, D. Batch 5 will create the same colors as the first batch.

Hope this helps!

8 0
3 years ago
Plz help me well mark brainliest if correct!
Charra [1.4K]

Answer:

diameter it goes throught the circle and as end points across

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Write 6.9 as a mixed number in simplest form.
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Ithw anawer is 6 9/10
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Tory plans to enter the path at mile marker 1 and hike for 4 miles. Tory is a beginner hiker, so he wants to choose the path tha
Sunny_sXe [5.5K]
If you need help visualizing this, you might draw vertical lines at distance = 1 and at distance = 5. Look at the points where those lines cross f(x). The vertical difference is perhaps 350 ft. Look at where the lines cross g(x). That vertical distance may be about 200 ft.

The vertical change from 1 to 5 is considerably less for g(x) than for f(x).

Tory should choose path ...
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