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3 years ago

If you give me answer that means you’re super smart

Mathematics

2 answers:

Zielflug [23.3K]3 years ago

7 0

Answer:

it is 160 if I remember correctly.

viktelen [127]3 years ago

4 0

Answer:

320

Step-by-step explanation:

- 2 × - 8 × - 2 × - 10

16 × - 2 × - 10

- 32 × - 10

320

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Three Cars for sale

Bumek [7]

Answer:

Car B.

Step-by-step explanation:

Car A : 12,380 = 12,400 --> diff : 20

Car B : 16,760 = 16,800 --> diff : 40

Car C : 14,580 = 14,600 --> diff : 20

7 0

3 years ago

Which is the answers

Jlenok [28]

Answers:

-|-9| = -9
-|9| = -9
the opposite of nine = -9

6 0

3 years ago

Pls answer this question

kupik [55]

Answer: 155y seconds

Step-by-step explanation:

Difference between 3y minutes and 25y seconds is written as

3y minutes- 25y seconds

First, Convert minutes to seconds = 60 seconds makes 1 minute, therefore, 3 minutes = 60 x 3= 180 seconds

Therefore 180y seconds - 25y seconds

= 155y seconds

I hope this helps, please mark as brainliest answer

4 0

3 years ago

Read 2 more answers

Simplify the following expression please solve correctly<br> −12^12 ^2 / 3^ 5 ^5

nikdorinn [45]

The simplified expression is $-\frac{48}{x^{3} y^{3}}$

Explanation:

The given expression is $\frac{-12x^{2}12y^{2} }{3x^{5} y^{5} }$

Let us simplify the expression.

Multiplying the numbers in the numerator, we have,

$\frac{-144 x^{2} y^{2}}{3 x^{5} y^{5}}$

Dividing the term 144 by 3, which results in 48.

Thus, we have,

$\frac{-48 x^{2} y^{2}}{x^{5} y^{5}}$

Applying the fraction rule $\frac{-a}{b}=-\frac{a}{b}$ , we get,

$-\frac{48 x^{2} y^{2}}{x^{5} y^{5}}$

Applying the exponent rule $\frac{x^{a}}{x^{b}}=\frac{1}{x^{b-a}}$ , we have,

$-\frac{48}{x^{3} y^{3}}$

Thus, the simplified expression is $-\frac{48}{x^{3} y^{3}}$

7 0

3 years ago

the sum of the perimeter of two squares is 40cm while the sum of the area is 58cm². Find length the side of the square.

tia_tia [17]

$a,\ b-the\ length\ of\ sides\\\\(1)\qquad4a+4b=40\qquad\text{divide both sides by 4}\\\\a+b=10\to a=10-b\\\\(2)\qquad a^2+b^2=58\\\\\text{Substitute}\ (1)\ \text{to}\ (2):\\\\(10-b)^2+b^2=58\qquad\text{Use}\ (x-y)^2=x^2-2xy+y^2\\\\10^2-2(10)(b)+b^2+b^2=58\\\\100-20b+2b^2=58\qquad\text{subtract 58 from both sides}\\\\42-20b+2b^2=0\\\\2b^2-20b+42=0\qquad\text{divide both sides by 2}\\\\b^2-10b+21=0\\\\b^2-7b-3b+21=0\\\\b(b-7)-3(b-7)=0\\\\(b-7)(b-3)=0\iff b-7=0\ \vee\ b-3=0\\\\\boxed{b=7\ \vee\ b=3}$

$\text{Put the values of b to }\ (1):\\\\for\ b=7\to a=10-7=3\\\\for\ b=3\to a=10-3=7\\\\Answer:\ \boxed{7cm\ and\ 3cm.}$

8 0

4 years ago