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2 years ago

Explain how solving -7y > 161 is different from solving 7y > - 161

2 answers:

8 0

Sample response: Both inequalities use the division property to isolate the variable, y. When you divide by a negative number, like –7, you must reverse the direction of the inequality sign. When you divide by a positive number, like 7, the inequality sign stays the same. The solution to the first inequality is y > -23, and the solution to the second inequality is y <>

IRISSAK [1]2 years ago

6 0

Step-by-step explanation:

-7y > 161

7y < 161

7÷7=1

161÷7=23

y=23

7y > -161

7÷7=1

-161÷7=-23

y=-23

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Last week, Lindsay earned $10 per hour plus a $60 bonus for good job performance. She spends StartFraction 1 Over 15 EndFraction

V125BC [204]

Answer:

StartFraction 1 Over 15 EndFraction left-parenthesis 10 h plus 60 right-parenthesis equals StartFraction 1 Over 10 Endfraction left-parenthesis 10 h right-parenthesis

Step-by-step explanation:

The desired equation equates 1/15 of the amount Lindsay earned with 1/10 the amount without the bonus:

StartFraction 1 Over 15 EndFraction left-parenthesis 10 h plus 60 right-parenthesis equals StartFraction 1 Over 10 Endfraction left-parenthesis 10 h right-parenthesis

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2 years ago

Read 2 more answers

Solve the system of linear equations using multiplication.

Anna71 [15]

Answer:

$(8,-1)$

Step-by-step explanation:

The given system is:

$3x+3y=21$

$6x+12y=36$

Since I prefer to use smaller numbers I'm going to divide both sides of the first equation by 3 and both sides of the equation equation by 6.

This gives me the system:

$x+y=7$

$x+2y=6$

We could solve the first equation for $x$ and replace the second $x$ with that.

Let's do that.

$x+y=7$

Subtract $y$ on both sides:

$x=7-y$

So we are replacing the second $x$ in the second equation with $(7-y)$ which gives us:

$(7-y)+2y=6$

$7-y+2y=6$

$7+y=6$

$y=6-7$

$y=-1$

Now recall the first equation we arranged so that $x$ was the subject. I'm referring to $x=7-y$ .

We can now find $x$ given that $y=-1$ using the equation $x=7-y$ .

Let's do that.

$x=7-y$ with $y=-1$ :

$x=7-(-1)$

$x=7+1$

$x=8$

So the solution is (8,-1).

We can check this point by plugging it into both equations.

If both equations render true for that point, then we have verify the solution.

Let's try it.

$3x+3y=21$ with $(x,y)=(8,-1)$ :

$3(8)+3(-1)=21$

$24+(-3)=21$

$21=21$ is a true equation so the "solution" looks promising still.

$6x+12y=36$ with $(x,y)=(8,-1)$ :

$6(8)+12(-1)=36$

$48+(-12)=36$

$36=36$ is also true so the solution has been verified since both equations render true for that point.

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2 years ago

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Congratulations you received a score of 3.4 on your annual review . Merit increases are given out starting at .5% at a score of

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1 year ago

Elliot has a total of 26 books. He has 12 more fiction books than nonfiction books. Let x represent the number of fiction books

Marta_Voda [28]

Given:

Elliot has a total of 26 books. He has 12 more fiction books than nonfiction books.

To Find:

A system of linear equations represents the situation.

Answer:

$x+y=26\\\\x=y+12$

Step-by-step explanation:

We are given that x represents the number of fiction books and y is the number of non-fiction books.

We are also given that the total number of books Elliot has is 26 which includes both fiction and non-fiction. So, we may write

$x+y=26$

Next, we are given that there are 12 more fiction books than non-fiction books. This means, the fiction books are more in number and so, we may write

$x=y+12$

So, the total system of equations can be represented as

$x+y=26\\\\x=y+12$

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2 years ago

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You have two circles, one with radius r and the other with radius R. You wish for the difference in the areas of these two circl

gtnhenbr [62]

Answer:

maximum difference of radii = $(r-R)=\frac{1}{2\pi ^{2}}$

Step-by-step explanation:

We know that area of circle is given by

$A=\pi \times (radius)^{2}$

For circle with radius 'r' we have

$A_{1}=\pi \times (r)^{2}$

For circle with radius 'R' we have

$A_{2}=\pi \times (R)^{2}$

Now according to given condition we have

$A_{1}-A_{2}\leq \frac{5}{\pi }$

$\Rightarrow \pi r^{2}-\pi R^{2}\leq \frac{5}{\pi }\\\\\Rightarrow (r^{2}-R^{2})\leq \frac{5}{\pi ^{2}}\\\\(r+R)(r-R)\leq \frac{5}{\pi ^{2}}\\\\\because (a^{2}-b^{2})=(a+b)(a-b)\\\\(r+R)=10(Given)\\\\\Rightarrow(r-R)\leq \frac{5}{10\pi ^{2}}\\\\\therefore (r-R)\leq\frac{1}{2\pi ^{2}}$

Thus maximum difference of radii = $(r-R)=\frac{1}{2\pi ^{2}}$

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2 years ago