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2 years ago

Solve by elimination x+y= 0 2x-y= 12

Mathematics

2 answers:

DIA [1.3K]2 years ago

8 0

Answer:

(x,y)= (4,-4)

Step-by-step explanation:

Thanks me later

dezoksy [38]2 years ago

5 0

Answer:

m=2

Step-by-step explanation:

I calculated it sorry

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Chance can wash the car in 2 hours, while Sandy can wash the same car in 4 hours. How many hours would it take for them to wash

aliya0001 [1]

Answer: First Option is correct.

Step-by-step explanation:

Since we have given that

Number of hours Chance can wash the car = 2 hours

Number of hours Sandy can wash the car = 4 hours

Work done by Chance per hour is given by

$\frac{1}{2}$

Work done by Sandy per hour is given by

$\frac{1}{4}$

Work done by altogether per hour is given by

$\frac{1}{2}+\frac{1}{4}\\\\=\frac{2+1}{4}\\\\=\frac{3}{4}$

So, Number of hours they take if they together wash the car is given by

$\frac{4}{3}$

Hence, First Option is correct.

6 0

2 years ago

Read 2 more answers

MIDPOINT & DISTANCE help

tamaranim1 [39]

Answer:

Step-by-step explanation:

The formula of a midpoint of (x₁, y₁) and (x₂, y₂):

$M\left(\dfrac{x_1+x_2}{2},\ \dfrac{y_1+y_2}{2}\right)$

We have two points (-3, 7) and (-9, 1).

Substitute:

$x=\dfrac{-3+(-9)}{2}=\dfrac{-12}{2}=-6\\\\y=\dfrac{7+1}{2}=\dfrac{8}{2}=4$

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2 years ago

Round each number to the place of the underlined digit the 4 is underlined in 324,650

Otrada [13]

Answer: my bad 324,000 isnt the answer. Its 325,000

Step-by-step explanation:

4 0

2 years ago

Using the properties of exponents and logarithms, find the value of x in 19 + 2 ln x = 25.

REY [17]

$19+2\ln x=25\ \ \ |-19\\\\2\ln x=6\ \ \ |:2\\\\\ln x=3\iff x=e^3$

$\text{Used de.finition of the logarithm:}\ \ \ \log_ab=c\iff a^c=b$

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2 years ago

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Determine whether the integral is convergent or divergent. ∫[infinity] 2 e^−1/x / x^2 dx : O Convergent O divergent If it is con

monitta

Let $f(x)=e^{-1/x}$ . Then $f'(x)=\frac1{x^2}e^{-1/x}>0$ for all $x\ge2$ , so $f$ is strictly increasing. As $x\to\infty$ , $e^{-1/x}\to e^0=1$ , so $f$ is bounded above by 1. This is to say,

$e^{-1/x}$

and the integral of $\frac1{x^2}$ converges over the same domain, so this integral must also converge by comparison.

We have, by setting $y=-\frac1x$ ,

$\displaystyle\int_2^\infty\frac{e^{-1/x}}{x^2}\,\mathrm dx=\int_{-1/2}^0e^y\,\mathrm dy=e^0-e^{-1/2}=1-\frac1{\sqrt e}$

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2 years ago

Solve by elimination x+y= 0 2x-y= 12​

Solve by elimination x+y= 0 2x-y= 12