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Maslowich
3 years ago
12

Which graph shows the points (−11

Mathematics
2 answers:
exis [7]3 years ago
8 0

Answer:

B

Step-by-step explanation:

got it right on edge

vodka [1.7K]3 years ago
4 0

Answer:

b

Step-by-step explanation: got it right on edge

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Solve y (1/2)y+6=(3/4)y–4
iren [92.7K]

Answer:

Y = 40

Step-by-step explanation:

Change 1/2 to 2/4

2/4 y + 6 = 3/4y - 4

subtract 2/4y

6 = 1/4y -4

add 4

10 = 1/4y

multiply by 4 to get fraction to 1y or y

y = 40

8 0
3 years ago
Read 2 more answers
Which is the graph of the linear inequality y &lt; 3x + 1?<br> 3<br> 1<br> 3<br> 4.<br> 5
Arlecino [84]

Answer:

Image below

Step-by-step explanation:

<em>Hey there!</em>

<em />

Well lets graph the given inequality,

y < 3x + 1

Look at the image below.

So the right graph should look like the image below.

<em>Hope this helps :)</em>

6 0
3 years ago
Original price is $50, discount is 15%, what is the sale price?
SashulF [63]

Answer: 42.50

Step-by-step explanation: sale price is the price at which something sells or is sold at after its price has been reduced

3 0
3 years ago
Read 2 more answers
1/3 *divide* 3/8 = m<br><br> help?
MariettaO [177]

Answer:

8/9

Step-by-step explanation:

To make it a fraction form answer, you multiply the dividend numerator by the divisor denominator to make a new numerator.

Furthermore, you multiply the dividend denominator by the divisor numerator to make a new denominator:

'

Thus, the answer to 1/3 divided by 3/8 in fraction form is: 8/9

5 0
3 years ago
Read 2 more answers
Explain how you could use the information that all circles are similar to show that the value of pi is a constant.
zhenek [66]

Answer:

The proof that πk(C1)=πk(C2) of course would just apply the similarity of polygons and the behavior of length and area for changes of scale. This argument does not assume a limit-based theory of length and area, because the theory of length and area for polygons in Euclidean geometry only requires dissections and rigid motions ("cut-and-paste equivalence" or equidecomposability). Any polygonal arc or region can be standardized to an interval or square by a finite number of (area and length preserving) cut-and-paste dissections. Numerical calculations involving the πk, such as ratios of particular lengths or areas, can be understood either as applying to equidecomposability classes of polygons, or the standardizations. In both interpretations, due to the similitude, the results will be the same for C1 and C2.

7 0
3 years ago
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