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liberstina [14]

4 years ago

15

1.70x+1.30y=57.90 5.60y+5.40y= 147.20

1 answer:

Annette [7]4 years ago

8 0

That's false. If I am understanding correctly.

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Find the value of y if B is between A and C, AB is 2y, BC is 6y, and AC is 48.

Len [333]

Answer:

7. C. 6

8. H. √34

9. A. (1, 3.5)

10. J. 10

Step-by-step explanation:

7. AB = 2y, BC = 6y, AC = 48

AB + BC = AC (segment addition theorem)

Substitute the above values into the equation

2y + 6y = 48

Solve for y

8y = 48

Divide both sides by 8

8y/8 = 48/8

y = 6

8. Distance between P(2, 8) and Q(5, 3):

$PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Let,

$P(2, 8) = (x_1, y_1)$

$Q(5, 3) = (x_2, y_2)$

$PQ = \sqrt{(5 - 2)^2 + (3 - 8)^2}$

$PQ = \sqrt{(3)^2 + (-5)^2}$

$PQ = \sqrt{9 + 25}$

$PQ = \sqrt{34}$

9. Midpoint (M) of segment LB, for L(8, 5) and B(-6, 2) is given as:

$M(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2})$

Let $L(8, 5) = (x_1, y_1)$

$B(-6, 2) = (x_2, y_2)$

Thus:

$M(\frac{8 + (-6)}{2}, \frac{5 + 2}{2})$

$M(\frac{2}{2}, \frac{7}{2})$

$M(1, 3.5)$

10. M = -10, N = -20

Distance between M and N, MN = |-20 - (-10)|

= |-20 + 10| = |-10|

MN = 10

5 0

4 years ago

<img src="https://tex.z-dn.net/?f=4%28a%20%2B%202%20%29%20-%202%20%28a%20-%203%29" id="TexFormula1" title="4(a + 2 ) - 2 (a - 3)

satela [25.4K]

Answer:

$\large\boxed{2a+14}$

Step-by-step explanation:

$4(a + 2 ) - 2 (a - 3)\qquad\text{use distributive property:}\ a(b+c)=ab+ac\\\\=4a+8-2a-(-6)\qquad\text{combine like terms}\\\\=(4a-2a)+(8+6)\\\\=2a+14$

5 0

3 years ago

Find the perimeter P of ▱JKLM with vertices J(−3,−2), K(−5,−5), L(1,−5), and M(3,−2). Round your answer to the nearest tenth, if

Bezzdna [24]

Given:

Vertices of JKLM are J(−3,−2), K(−5,−5), L(1,−5), and M(3,−2).

To find:

The perimeter P of a parallelogram JKLM.

Solution:

Distance formula:

$D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Using distance formula, we get

$JK=\sqrt{\left(-5-\left(-3\right)\right)^2+\left(-5-\left(-2\right)\right)^2}$

$JK=\sqrt{\left(-5+3\right)^2+\left(-5+2\right)^2}$

$JK=\sqrt{\left(-2\right)^2+\left(-3\right)^2}$

$JK=\sqrt{4+9}$

$JK=\sqrt{13}$

Similarly,

$KL=\sqrt{\left(1-\left(-5\right)\right)^2+\left(-5-\left(-5\right)\right)^2}=6$

$LM=\sqrt{\left(3-1\right)^2+\left(-2-\left(-5\right)\right)^2}=\sqrt{13}$

$JM=\sqrt{\left(3-\left(-3\right)\right)^2+\left(-2-\left(-2\right)\right)^2}=6$

Now, perimeter P of ▱JKLM is

$P=JK+KL+LM+JM$

$P=\sqrt{13}+6+\sqrt{13}+6$

$P=2\sqrt{13}+12$

$P=2(3.61)+12$

$P=7.22+12$

$P=19.22$

$P\approx 19.2$

Therefore, the perimeter P of ▱JKLM is 19.2 units.

3 0

3 years ago

What is the value of x if a rectangle has a perimeter of 50, a length of x + 5, and width of 11?

Iteru [2.4K]

The perimeter of a polygon is the sum of the lengths of all its sides. A rectangle has 4 sides, and opposite sides are congruent, so 2 sides have length x + 5, and 2 sides have length 11. Add all the lengths and set the sum equal to 50. Then solve the equation for x.

x + 5 + x + 5 + 11 + 11 = 50

2x + 32 = 50

2x = 18

x = 9

Answer: x = 9

6 0

4 years ago

-4(-3n - 8 ) = 10n + 20<br><br> n =

IrinaK [193]

Answer: n = -6

Step-by-step explanation:

−4(−3n−8)=10n+20

n = -6

7 0

3 years ago