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4 years ago

Ajuda por favor!! Help please!

Mathematics

1 answer:

adelina 88 [10]4 years ago

5 0

Answer:Me no speak espanol

Step-by-step explanation:

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X=(x-(x - y)); use x = 9, and y= 1

Olegator [25]

Answer:

I think the answer for this question is x=1

x=(9(9-1)

Step-by-step explanation:

7 0

4 years ago

What would happen if a nation had an unlimited supply of a certain resource?

Lisa [10]

Answer:

the country will lack that particular resource and the country will lack the necessary products from that resource.

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3 years ago

Select 3 expressions that have a sum or difference of 3 /4 .

cupoosta [38]

Answer:

The Three expressions that have a sum or difference of 3 /4 .

B . 11/12 − 1/6

C. 3/5 + 3/20

E . 2/3 + 1/12

Step-by-step explanation:

The Three expressions that have a sum or difference of 3 /4 .

are

B . 11/12 − 1/6

$\dfrac{11}{12}-\dfrac{1}{6}=\dfrac{11}{12}-\dfrac{1\times 2}{6\times 2}\\\\\dfrac{11}{12}-\dfrac{1}{6}=\dfrac{11}{12}-\dfrac{2}{12}=\dfrac{11-2}{12}=\dfrac{9}{12}=\dfrac{3}{4}$

Therefore,

$\dfrac{11}{12}-\dfrac{1}{6}=\dfrac{3}{4}$

C. 3/5 + 3/20

$\dfrac{3}{5}+\dfrac{3}{20}=\dfrac{3\times 4}{5\times 4}+\dfrac{3}{20}\\\\\dfrac{3}{5}+\dfrac{3}{20}=\dfrac{12}{20}+\dfrac{3}{20}=\dfrac{15}{20}=\dfrac{3}{4}$

Therefore,

$\dfrac{3}{5}+\dfrac{3}{20}=\dfrac{3}{4}$

E . 2/3 + 1/12

$\dfrac{2}{3}+\dfrac{1}{12}=\dfrac{2\times 4}{3\times 4}+\dfrac{1}{12}\\\\\dfrac{3}{5}+\dfrac{3}{20}=\dfrac{8+1}{12}=\dfrac{9}{12}=\dfrac{3}{4}$

Therefore,

$\dfrac{2}{3}+\dfrac{1}{12}=\dfrac{3}{4}$

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3 years ago

You want to wrap a gift box that is 3 in. long, 5 in. wide and 10 in. high. How much gift wrap will you need to cover the entire

goldenfox [79]

Answer:

150

Step-by-step explanation:

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3 years ago

From the sum of 3a^2−ab−2b^2 and 2a^2+5ab−3b^2 subtract a^2−3ab−4b^2

Lubov Fominskaja [6]

$(3a^2-ab-2b^2) + (2a^2+5ab-3b^2) - (a^2-3ab-4b^2)$

First we distribute the negative throughout the last polynomial:

$(3a^2-ab-2b^2) + (2a^2+5ab-3b^2) + (-a^2+3ab+4b^2)$

Combine like terms:

$4a^2 + 7ab - b^2$

3 0

3 years ago