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3 years ago

Solve 3(y-2)=6(y-1)-3y

Mathematics

1 answer:

ASHA 777 [7]3 years ago

3 0

Answer:

All real numbers are solutions.

Step-by-step explanation:

1. First you want to distribute what is inside the parenthesis. You would want to do 3 × y + 3 × -2 = 6 × y + 6 × -1 -3y.

2. Now that the equation has been distributed, you want to simplify it. 3 × y is equal to 3y, and 3 × -2 is equal to -6. So, 3y + -6 = 6 × y + 6 × -1 -3y.

3. Now you want to simplify the right side of the equation. So, 6 times y is equal to 6y, and 6 × -1 is equal to -6. Now the new equation is 3y + -6 =6y + -6-3y.

4. Also in an equation, if there is a + and - sign together like this +-, it would equal to -. So, 3y + -6 = 6y + -6-3y, is equal to 3y - 6 = 6y -6-3y.

5. Now you want to combine like terms on the right side of the equation. 3y - 6 = 3y -6 because 6y-3y is equal to 3y.

6. Now the rest of the equation has to be combined using like terms. Start by moving the -3y fromthe right side of the equation to the 3y on the left side of the equation. -3y plus 3y is equal to zero so it cancels out. The 3y on the left side plus another 3y is equal to 6y.

7. Now the equation is 6y - 6 = 6y - 6

8. You can add 6 to the -6 on both sides of the equation which would cancel out on both sides leaving you with an equation of 6y = 6y.

9. Divide both sides by 6 to get y = y.

10. This equation means that all real numbers are solutions to this equation. In other words, no matter what real number you use in this equation, it would come out to be true. For example, if y = 3, 3 = 3, so the equation is true.

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Lina20 [59]

Step 1: We make the assumption that 498 is 100% since it is our output value.

Step 2: We next represent the value we seek with $x$x.

Step 3: From step 1, it follows that $100\%=498$100%=498.

Step 4: In the same vein, $x\%=4$x%=4.

Step 5: This gives us a pair of simple equations:

$100\%=498(1)$100%=498(1).

$x\%=4(2)$x%=4(2).

Step 6: By simply dividing equation 1 by equation 2 and taking note of the fact that both the LHS

(left hand side) of both equations have the same unit (%); we have

$\frac{100\%}{x\%}=\frac{498}{4}$

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x%=

498

Step 7: Taking the inverse (or reciprocal) of both sides yields

$\frac{x\%}{100\%}=\frac{4}{498}$

100%=

498

$\Rightarrow x=0.8\%$⇒x=0.8%

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2 years ago

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leonid [27]

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3 years ago

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Find the distance from the origin to the graph of 7x+9y+11=0

Cerrena [4.2K]

One way to do it is with calculus. The distance between any point $(x,y)=\left(x,-\dfrac{7x+11}9\right)$ on the line to the origin is given by

$d(x)=\sqrt{x^2+\left(-\dfrac{7x+11}9\right)^2}=\dfrac{\sqrt{130x^2+154x+121}}9$

Now, both $d(x)$ and $d(x)^2$ attain their respective extrema at the same critical points, so we can work with the latter and apply the derivative test to that.

$d(x)^2=\dfrac{130x^2+154x+121}{81}\implies\dfrac{\mathrm dd(x)^2}{\mathrm dx}=\dfrac{260}{81}x+\dfrac{154}{81}$

Solving for $(d(x)^2)'=0$ , you find a critical point of $x=-\dfrac{77}{130}$ .

Next, check the concavity of the squared distance to verify that a minimum occurs at this value. If the second derivative is positive, then the critical point is the site of a minimum.

You have

$\dfrac{\mathrm d^2d(x)^2}{\mathrm dx^2}=\dfrac{260}{81}>0$

so indeed, a minimum occurs at $x=-\dfrac{77}{130}$ .

The minimum distance is then

$d\left(-\dfrac{77}{130}\right)=\dfrac{11}{\sqrt{130}}$

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3 years ago

HELPPPP What is MB 48 96 110 55

Lady_Fox [76]

Minor Arc AC = 110 degrees (given)
As indicated in the hints,
OA=OC => OAC=OCA=(180-110)/2 = 35
Similarly,
OAB=OBA=(48-35)=13
=>
OBC=OCB=(180-35-35-13-13)/2=(180-96)/2=84/2=42
=>
Angle B = 13+42 = 55 degrees

Check:
55+55+35+35+13+13 = 180 degrees (angles of a triangle) ok.

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