Which digit represents "hundreds" in the number 8732?

Find the missing coefficient.<br><br> -<br> y2 − [-5y − y(-7y − 9)] − [-y (15y + 4)] = 0

maria [59]

There is only one solution in the given equation -y2 − [-5y − y(-7y − 9)] − [-y (15y + 4)] = 0. In solving this problem, apply first PEMDAS (parenthesis, exponents,multiplication, division, addition, subtraction). Then equation will transform into -y2+5y-7y2-9y+15y2+4y=0. Combine terms with same power and achieve 7y2=0. Divide both sides with 7 and perform square root of zero. Since the root is zero, we have one solution of the given equation which is y=0.

5 0

4 years ago

Read 2 more answers

Quadrilateral ABCD is dilated by a scale factor of 1 over 2 centered around (2, 2). Which statement is true about the dilation?

Musya8 [376]

Answer:

Option (1)

1) Segment B'D' will run through (2, 2) and will be shorter than segment BD.

Step-by-step explanation:

The rest of the question is the attached figure (1).

The statement options are:

1) Segment B'D' will run through (2, 2) and will be shorter than segment BD.

2) Segment B'D' will run through (2, 2) and will be longer than segment BD.

3) Segment B'D' will parallel to segment BD and will be shorter than segment BD.

4) Segment B'D' will be parallel to segment BD and will be longer than segment BD.

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See the attched figure (2), which represents Quadrilateral ABCD and the image A'B'C'D'

As shown:

The quadrilateral ABCD with vertices A(1,2), B(2,3), C(4,2) and D(2,1).

Quadrilateral ABCD is dilated by a scale factor of 1 over 2 centered around (2, 2).

So,

The quadrilateral A'B'C'D' will be with vertices:

A'= (1.5,2), B'= (2,2.5), C'= (3,2), D'= (2,1.5)

Comparing the options with the figure (2):

So, the answer is option (1)

1) Segment B'D' will run through (2, 2) and will be shorter than segment BD.

6 0

3 years ago

HELP ME I NEED HELP!!!!

eduard

Answer:

D

Step-by-step explanation:

Since they are the same variables

7 0

3 years ago

Mr. Thomas drove 75 miles in May. He drove 6 times as many miles in July as he did in May. He

harkovskaia [24]

Answer:

1800 miles

Step-by-step explanation:

No. of miles driven by Mr. Thomas in May = 75

It is given that miles driven in July is 6 times of miles driven by Mr. Thomas in May(75 miles).

Thus

No. of miles driven by Mr. Thomas in July = 6 * No. of miles driven by Mr. Thomas in May = 6*75 = 450 miles.

__________________________________________________

Another condition given that miles driven in June is 4 times of miles driven by Mr. Thomas in July(450miles as calculated above).

Thus

No. of miles driven by Mr. Thomas in June = 4 * No. of miles driven by Mr. Thomas in July = 4* 450 miles = 1800 miles.

No. of miles driven by Mr. Thomas in June is 1800 miles.

8 0

3 years ago

Need help in number 12 and 13 PLEASEE!! I don’t get it and I start school the day after tomorrow

Natasha_Volkova [10]

Answer:

The explanations for the graphs are provided down below. Please let me know if you have any questions about my answer.

12 and 13 as written on the worksheet is right.

Step-by-step explanation:

12) The answer given is correct.

The relation between x and y is given as:

$y=\frac{x^2}{2}-3$ with $x \in \{-4,-2,0,2}$ .

I replaced the word domain with x since the domain is the set of x's for which the relation exists.

We are going to replace x with each of the x's given to see what y corresponds to each.

Let's begin with x=-4.

$y=\frac{x^2}{2}-3$ with $x=-4$ :

$y=\frac{(-4)^2}{2}-3$

$y=\frac{16}{2}-3$

$y=8-3$

$y=5$ .

So (-4,5) is an ordered pair that should be on our graph.

To find this point you move left 4 from origin then up 5. Now you put a dot where you have landed. Your graph does show this point.

Moving on.

Let's do the next x: x=-2.

$y=\frac{x^2}{2}-3$ with $x=-2$ :

$y=\frac{(-2)^2}{2}-3$

$y=\frac{4}{2}-3$

$y=2-3$

$y=-1$ .

So (-2,-1) is an ordered pair that should be on our graph.

To find this point you move left 2 from origin and then down 1. Now you put a dot where you have landed. Your graph shows this point as well.

Now x=0.

$y=\frac{x^2}{2}-3$ with $x=0$ :

$y=\frac{0^2}{2}-3$

$y=\frac{0}{2}-3$

$y=0-3$

$y=-3$

So (0,-3) is an ordered pair that should be on our graph.

To find this point you move left and right none and down 3. Now you put a dot where you have landed. Your graph shows this point.

Now the last point will be at x=2.

$y=\frac{x^2}{2}-3$ with $x=2$

$y=\frac{2^2}{2}-3$

$y=\frac{4}{2}-3$

$y=2-3$

$y=-1$ .

So (2,-1) is an ordered pair that should be on our graph.

To find this point you move 2 units right from the origin and then down 1 unit. Now put a dot where you landed. The graph shows this point as well.

13) The answer given is correct.

$g(x)=|x|$ is the parent function and makes like a V shaped graph where it's vertex is at (0,0).

If we want to move this graph right 3 it becomes:

$m(x)=|x-3| \text{ or } m(x)=|(-1)(-x+3)|=|-1||-x+3|=1|-x+3|=|-x+3|=|3-x|$ .

If you move that up once it becomes:

$n(x)=|x-3|+1$ or $n(x)=|3-x|+1$ which is the curve given.

If you don't know about transformations you can choose a few points to plug in to see what's going on with the graph.

Let's choose x=-5,-3,-1,0,1,3,5.

x=-5

f(-5)=|3--5|+1=|3+5|+1=|8|+1=8+1=9.

There is no room for (-5,9) on our graph but if you extended the left hand side of the absolute value function there you would see that (-5,9) is reached.

x=-3

f(-3)=|3--3|+1=|3+3|+1=|6|+1=6+1=7.

(-3,7) should be a point on the graph. Same thing for this point as (-5,9).

x=-1

f(-1)=|3--1|+1=|3+1|+1=|4|+1=4+1=5.

(-1,5) is located on the graph.

x=0

f(0)=|3-0|+1=|3|+1=3+1=4.

(0,4) is also located on the graph.

x=1

f(1)=|3-1|+1=|2|+1=2+1=3.

(1,3) is located on the graph.

x=3

f(3)=|3-3|+1=|0|+1=0+1=1.

(3,1) is located on the graph.

x=5

f(5)=|3-5|+1=|-2|+1=2+1=3.

(5,3) is located on the graph.

Now if we weren't given the graph already:

I would plot the points I found which were:

(-5,9)

(-3,7)

(-1,5)

(0,4)

(1,3)

(3,1)

(5,3)

We should get a basic idea of what the function looks like from these points.

I will graph them. You will have to connect these points because the domain isn't discrete like number 12 is. That is they didn't list out elements for the domain.

I'm going to graph one more point after x=5.

How about x=7?

$f(7)=|3-7|+1=|-4|+1=4+1=5$

So (7,5) is also a point on the graph.

You should see that the blue points are following the red path I made there.

8 0

3 years ago