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4 years ago

4.5w=5.1w-30, how do I solve it

2 answers:

5 0

First you have to get the same variables on one side so you’d subtract 5.1w from both sides, making the equation -0.6w = -30

then you want to single out the variable, diving -0.6 on both sides making the new equation and the answer w = 50

hope this helps!

mina [271]4 years ago

5 0

Hey there!

Let's subtract by $5.1w$ on EACH of your sides that we're working with

LIke: $4,5w - 5.1w \\ = 5.1w - 30 - 5.1w$

We get: $-0.6w = -30$

Now, let's divide by $-0.6$ on EACH of your sides

LIke: $\frac{-0.6w}{-0.6} 
\\ \\ = \frac{-30}{-0.6}$

$Answer: w = 50$

Good luck on your assignment and enjoy your day!

~ $MeIsKaitlyn:)$

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attashe74 [19]

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Step-by-step explanation:

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8 0

2 years ago

Solve the system of equations by elimination. 2x+8y=10 -4x-9y=-13 x = y =

Reptile [31]

So use cancelation
multiply first equation by 2
2x+8y=10
times 2
4x+16y=20
now add

4x+16y=20
-4x-9y=-13 +
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7y=7
divide by 7
y=1

subsittue

2x+8y=10
2x+8(1)=10
2x+8=10
subtract 8 from both sides
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divide by 2
x=1

x=1
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8 0

3 years ago

Evaluate the integral using the indicated trigonometric substitution. (use c for the constant of integration.) x^3 / sqrt x^2 +

slava [35]

$\displaystyle\int\frac{x^3}{\sqrt{x^2+49}}\,\mathrm dx$

Taking $x=7\tan\theta$ gives $\mathrm dx=7\sec^2\theta\,\mathrm d\theta$ , so that the integral becomes

$\displaystyle\int\frac{(7\tan\theta)^3}{\sqrt{(7\tan\theta)^2+49}}(7\sec^2\theta)\,\mathrm d\theta$
$=\displaystyle7^4\int\frac{\tan^3\theta\sec^3\theta}{\sqrt{49\tan^2\theta+49}}\,\mathrm d\theta$
$=\displaystyle7^3\int\frac{\tan^3\theta\sec^3\theta}{\sqrt{\tan^2\theta+1}}\,\mathrm d\theta$
$=\displaystyle7^3\int\frac{\tan^3\theta\sec^3\theta}{\sqrt{\sec^2\theta}}\,\mathrm d\theta$
$=\displaystyle7^3\int\frac{\tan^3\theta\sec^3\theta}{|\sec\theta|}\,\mathrm d\theta$

When $\sec\theta>0$ , we have

$=\displaystyle7^3\int\frac{\tan^3\theta\sec^3\theta}{\sec\theta}\,\mathrm d\theta$
$=\displaystyle7^3\int\tan^3\theta\sec^2\theta\,\mathrm d\theta$

and from here we can substitute $u=\tan\theta$ to proceed from here.

Quick note: When we set $x=7\tan\theta$ , we are implicitly enforcing $-\dfrac\pi2$ just so that the substitution can be undone later via $\theta=\tan^{-1}\dfrac x7$ . But note that over this domain, we automatically guarantee that $\sec\theta>0$ , so the absolute value bars can be dropped immediately.

6 0

3 years ago

3 less than the product of 5 and x?

amm1812

Answer:

5x - 3

The product of 5 and x, minus 3

4 0

3 years ago

A ladder 20 feet long leans against a building forming and angle of 46 degrees with the level ground. To the nearest foot how fa

sleet_krkn [62]

Answer:

The distance of the foot of the ladder to the building is 14 ft.

Step-by-step explanation:

The length of ladder = 20 ft

Angle formed by ladder with level ground, θ = 46

We are required to find out the distance of the foot of the ladder from the building

The above question can be found out by using trigonometric relations as follows;

$Cos\theta = \frac{Adjacent\, side \, to\, angle}{Hypothenus\, side \, of\, triangle}$

The adjacent side of the right triangle formed by the ladder the building and the ground is the distance of the foot of the ladder from the building

The hypotenuse side is the length of the ladder = 20 ft

Therefore;

Adjacent side of triangle = Hypotenuse × cosθ

∴ Distance of the foot of the ladder from the building = Hypotenuse × cosθ

Distance of the foot of the ladder from the building = 20 ft × cos(56)

Distance of the foot of the ladder from the building = 13.893 ft

To the nearest foot, the distance of the foot of the ladder to the building = 14 ft.

7 0

3 years ago