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2 years ago

And it's elimination using addition and subtraction x-y=1 x+y=3

2 answers:

7 0

Get one x+y=3 to y=x+3 and plug it into x-y=1 and you get x-(x+3)=1. Destitute the x and solve. Once you get what x equals plug it to to either equation for the value of x and solve for y

nirvana33 [79]2 years ago

6 0

X-y=1. ....> x=1+y
(1+y)+y =3
1+y+y =3
1+2y =3
2y=3-1=2 ....>y =2/2=1
Y=1
X-y=1
X-1=1. ....>x=1+1 =2

X=2. Y=1

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MaRussiya [10]

Just divide it and you will get the answer

5 0

2 years ago

If b + 2 = 4.6 Then what is b?

Morgarella [4.7K]

The answer is 2.6 because 2.6+2=4.6

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2 years ago

Read 2 more answers

Miguel collected aluminum cans for recycling. He collected a total of 150 cans. If 58% of the cans he collected were soda cans h

Romashka-Z-Leto [24]

Given:

• Total number of cans collected = 150

• Percent of cans that were soda = 58%

Let's find the number of other cans he collected.

To find the number of other cans, since the percent of soda is 58%, let's find the pecent of other cans.

Percent of other cans = 100% - 58% = 42%

The percent of other cans collected was 42%.

Now, to find the number of other cans collected, let's find 42% of the total number of cans collected 150.

We have:

$42\text{ \% }\ast150\Longrightarrow0.42\times150=63$

Therefore, the number of other cans collected is 63 cans.

ANSWER:

63 cans

4 0

2 months ago

Find the value of x

maks197457 [2]

Answer:

Step-by-step explanation:

∠S=180-(54+62)=180-116=64°

using sin formula

$\frac{x}{sin~64} =\frac{7}{sin~62} \\x=\frac{7~sin~64}{sin~62} \approx 7.1256 \approx 7.13$

6 0

2 years ago

1.) Find the length of the arc of the graph x^4 = y^6 from x = 1 to x = 8.

xxTIMURxx [149]

First, rewrite the equation so that y is a function of x :

$x^4 = y^6 \implies \left(x^4\right)^{1/6} = \left(y^6\right)^{1/6} \implies x^{4/6} = y^{6/6} \implies y = x^{2/3}$

(If you were to plot the actual curve, you would have both $y=x^{2/3}$ and $y=-x^{2/3}$ , but one curve is a reflection of the other, so the arc length for 1 ≤ x ≤ 8 would be the same on both curves. It doesn't matter which "half-curve" you choose to work with.)

The arc length is then given by the definite integral,

$\displaystyle \int_1^8 \sqrt{1 + \left(\frac{\mathrm dy}{\mathrm dx}\right)^2}\,\mathrm dx$

We have

$y = x^{2/3} \implies \dfrac{\mathrm dy}{\mathrm dx} = \dfrac23x^{-1/3} \implies \left(\dfrac{\mathrm dy}{\mathrm dx}\right)^2 = \dfrac49x^{-2/3}$

Then in the integral,

$\displaystyle \int_1^8 \sqrt{1 + \frac49x^{-2/3}}\,\mathrm dx = \int_1^8 \sqrt{\frac49x^{-2/3}}\sqrt{\frac94x^{2/3}+1}\,\mathrm dx = \int_1^8 \frac23x^{-1/3} \sqrt{\frac94x^{2/3}+1}\,\mathrm dx$

Substitute

$u = \dfrac94x^{2/3}+1 \text{ and } \mathrm du = \dfrac{18}{12}x^{-1/3}\,\mathrm dx = \dfrac32x^{-1/3}\,\mathrm dx$

This transforms the integral to

$\displaystyle \frac49 \int_{13/4}^{10} \sqrt{u}\,\mathrm du$

and computing it is trivial:

$\displaystyle \frac49 \int_{13/4}^{10} u^{1/2} \,\mathrm du = \frac49\cdot\frac23 u^{3/2}\bigg|_{13/4}^{10} = \frac8{27} \left(10^{3/2} - \left(\frac{13}4\right)^{3/2}\right)$

We can simplify this further to

$\displaystyle \frac8{27} \left(10\sqrt{10} - \frac{13\sqrt{13}}8\right) = \boxed{\frac{80\sqrt{10}-13\sqrt{13}}{27}}$

7 0

2 years ago