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lakkis [162]
3 years ago
10

4) -4y= -x + 4 6) y + 3x = 3

Mathematics
1 answer:
wel3 years ago
4 0

Answer:

x=16/13

y= -9/13

Step-by-step explanation:

A:         -4y=-x+4

B:          y+3x=3

4B:       4y+12x=12

A+4B:   12x= -x+16

----->    13x=16

               x=16/13

-4y= -16/13+4

4y=16/13-4

y=4/13-1= -9/13

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What is the square units of 14.8099
VashaNatasha [74]
We grapht hem

we see that the base is from (-2,2) and (6,2), 8 units long
the height will be from y=2 to y=8 or 6 units
so then, the area=1/2 times base times height=1/2 times 8 times 6=24 square units
5 0
3 years ago
Which statement best describes her claim?
Norma-Jean [14]

Answer:

B: She is incorrect because the segments do not have the same length.

Step-by-step explanation:

3 0
3 years ago
After triangle ACE is dilated by a factor of 6, it has an area of 180 square inches. What was its area before dilation?
Agata [3.3K]

Answer:

The area of triangle ACE before dilation is:  30 square inches

Step-by-step explanation:

We know that when an object is dilated by a scale factor, it gets reduced, stretched, or remains the same, depending upon the value of the scale factor.

  • If the scale factor > 1, the image is enlarged
  • If the scale factor is between 0 and 1, it gets shrunk
  • If the scale factor = 1, the object and the image are congruent

The length of the image can be determined by multiplying the length of the original point by a certain scale factor.

Given that after triangle ACE is dilated by a factor of 6, it has an area of 180 square inches.

Thus, it means the dilated area is 6 times the area before dilation.

Let x be the pre-dilation area

As the dilated area is 6 times the area before dilation, so

6x = 180

divide both sides by 6

6x/6 = 180/6

x = 30 square inches.

Therefore, the area of triangle ACE before dilation is:  30 square inches

4 0
3 years ago
Please help with this problem :(
Lilit [14]

36 - 24

= 12 + 16

= 28 - 10 2/3

<h3>= 17 1/3</h3><h3 />

Best of Luck to you.

8 0
3 years ago
For the following telescoping series, find a formula for the nth term of the sequence of partial sums
gtnhenbr [62]

I'm guessing the sum is supposed to be

\displaystyle\sum_{k=1}^\infty\frac{10}{(5k-1)(5k+4)}

Split the summand into partial fractions:

\dfrac1{(5k-1)(5k+4)}=\dfrac a{5k-1}+\dfrac b{5k+4}

1=a(5k+4)+b(5k-1)

If k=-\frac45, then

1=b(-4-1)\implies b=-\frac15

If k=\frac15, then

1=a(1+4)\implies a=\frac15

This means

\dfrac{10}{(5k-1)(5k+4)}=\dfrac2{5k-1}-\dfrac2{5k+4}

Consider the nth partial sum of the series:

S_n=2\left(\dfrac14-\dfrac19\right)+2\left(\dfrac19-\dfrac1{14}\right)+2\left(\dfrac1{14}-\dfrac1{19}\right)+\cdots+2\left(\dfrac1{5n-1}-\dfrac1{5n+4}\right)

The sum telescopes so that

S_n=\dfrac2{14}-\dfrac2{5n+4}

and as n\to\infty, the second term vanishes and leaves us with

\displaystyle\sum_{k=1}^\infty\frac{10}{(5k-1)(5k+4)}=\lim_{n\to\infty}S_n=\frac17

7 0
3 years ago
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