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2 years ago

Draw a rectangle with an area of 42 square units

1 answer:

6 0

We know, Area = Length * Width

So, with area 42, dimensions could be: 1 * 42
2 * 21
3 * 14
6 * 7

Just draw the lines with that measures, two pair of two equal opposite lines.

Hope this helps!

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Imagine the United States only had two postage stamps, a 3 cent stamp and a 7 cent stamp. If you can put any number of stamps on

cricket20 [7]

Answer:

not possible: 1, 2, 4, 5, 8, 11
possible: all others

Step-by-step explanation:

Obviously, all multiples of 3 and 7 can be paid.

1 cent can be added by adding a 7-cent stamp and taking away two 3-cent stamps. 2 cents can be added by adding three 3-cent stamps and taking away one 7-cent stamp. Hence any value more than 7 +2(3) = 13 cents can always be accommodated.

Amounts that are impossible:

1, 2, 4, 5, 8, 11 cents

All other positive integer amounts are possible.

7 0

3 years ago

Is the relation a function?

Margaret [11]

No; a domain value has two range values.

x = -2 then y = 1 and 2

they would form a vertical line, which tells us that it's not a function

8 0

3 years ago

Bess [88]

Answer:

-7+12 will also give u 5 with a positive sign

8 0

2 years ago

76 19 in ratio simplest form

Nata [24]

Putting the two numbers together in fraction form of a ratio, we have 76/19. We know that something multiplied by 19 will equal to 76.

When we try to multiply 4 and 19 together, we have our nice answer 76. We then replace out values in the original ratio, in which we get the final answer of 4:1, 4/1, or 4 to 1.

8 0

2 years ago

Will mark Brainlest hellpppp

Anna007 [38]

$h(-3) - h( - 2) = - \frac{5}{ 8} \\ \\$

Step-by-step explanation:

$\boxed{h(x) = \frac{2 {x}^{2} - x + 1}{3x - 2}}$

$h(2) = \frac{2 {(2)}^{2} - (2) + 1}{3(2) - 2}$

$h(2) = \frac{2 {(4)} - 2 + 1}{6 - 2}$

$h(2) = \frac{ 8 - 1}{4}$

$h(2) = \frac{7}{4}$

$h( - 3) = \frac{2 {( - 3)}^{2} - ( - 3) + 1}{3( - 3) - 2} \\ h( - 3) = \frac{2 (9) + 3 + 1}{ - 9- 2} \\ h( - 3) = \frac{18 + 4}{ - 11} \\ h( - 3) = \frac{22}{ - 11} \\ h(-3) = -2$

$h( - 2) = \frac{2 {( - 2)}^{2} - ( - 2) + 1}{3( - 2) - 2} \\ h( - 2) = \frac{2 (4) + 2+ 1}{- 6 - 2} \\ h( - 2) = \frac{8 + 3}{- 8} \\ h( - 2) = \frac{11}{- 8}$

$h(3) - h( - 2) = -2 - \frac{11}{- 8} \\ h(3) - h( - 2) =-2 + \frac{11}{ 8} \\ h(3) - h( - 2) = \frac{-2}{1}+ \frac{11}{ 8} \\ h(3) - h( - 2) = \frac{-16}{8} + \frac{11}{ 8} \\ h(3) - h( - 2) = \frac{-16+11}{8} \\ h(3) - h( - 2) = \frac{-5}{8} \\ - \frac{5}{ 8}$

5 0

2 years ago