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3 years ago

Identify the like terms.

Mathematics

1 answer:

lesya [120]3 years ago

3 0

C! Like terms refer to the variables that are the same, rather than the coefficients (the numbers).

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PLEASE HEWP EASY SCIENCE BRIANLIST U CAN BE IF CORRECT

erik [133]

Answer:

65kg

Step-by-step explanation:

4 0

3 years ago

ANYONE KNOW THE ANSWER TO THIS??

mojhsa [17]

Answer:

x = 14 is your answer

Step-by-step explanation:

76 = 7x-22, so to figure it out, you first move -22 to the other side and get

76+22 = 7x.

add them together and you get 98 = 7x

7x would be your equation, so divide 98 by 7 and you get your solution

x = 14 is your answer

3 0

3 years ago

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A line through which two points would be parallel to a line with a slope of 3/4?

Mumz [18]

Answer:

F. (0, 5) and (-4, 2)

Step-by-step explanation:

We need to calculate the slope of each of the given sets of points until we find the set associated with a slope of 3/4:

F. (0,5) and (-4,2) As we go from (-4, 2) to (0,5), x increases by 4 and y increases by 3, so the slope is m = rise / run = 3/4. This is the line with slope 3/4.

5 0

3 years ago

Given the center of the circle (-3,4) and a point on the circle (-6,2), (10,4) is on the circle

Anastasy [175]

Answer:

Part 1) False

Part 2) False

Step-by-step explanation:

we know that

The equation of the circle in standard form is equal to

$(x-h)^{2} +(y-k)^{2}=r^{2}$

where

(h,k) is the center and r is the radius

In this problem the distance between the center and a point on the circle is equal to the radius

The formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

Part 1) given the center of the circle (-3,4) and a point on the circle (-6,2), (10,4) is on the circle.

true or false

substitute the center of the circle in the equation in standard form

$(x+3)^{2} +(y-4)^{2}=r^{2}$

Find the distance (radius) between the center (-3,4) and (-6,2)

substitute in the formula of distance

$r=\sqrt{(2-4)^{2}+(-6+3)^{2}}$

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

$r=\sqrt{13}\ units$

The equation of the circle is equal to

$(x+3)^{2} +(y-4)^{2}=(\sqrt{13}){2}$

$(x+3)^{2} +(y-4)^{2}=13$

Verify if the point (10,4) is on the circle

we know that

If a ordered pair is on the circle, then the ordered pair must satisfy the equation of the circle

For x=10,y=4

substitute

$(10+3)^{2} +(4-4)^{2}=13$

$(13)^{2} +(0)^{2}=13$

$169=13$ -----> is not true

therefore

The point is not on the circle

The statement is false

Part 2) given the center of the circle (1,3) and a point on the circle (2,6), (11,5) is on the circle.

true or false

substitute the center of the circle in the equation in standard form

$(x-1)^{2} +(y-3)^{2}=r^{2}$

Find the distance (radius) between the center (1,3) and (2,6)

substitute in the formula of distance

$r=\sqrt{(6-3)^{2}+(2-1)^{2}}$

$r=\sqrt{(3)^{2}+(1)^{2}}$

$r=\sqrt{10}\ units$

The equation of the circle is equal to

$(x-1)^{2} +(y-3)^{2}=(\sqrt{10}){2}$

$(x-1)^{2} +(y-3)^{2}=10$

Verify if the point (11,5) is on the circle

we know that

If a ordered pair is on the circle, then the ordered pair must satisfy the equation of the circle

For x=11,y=5

substitute

$(11-1)^{2} +(5-3)^{2}=10$

$(10)^{2} +(2)^{2}=10$

$104=10$ -----> is not true

therefore

The point is not on the circle

The statement is false

7 0

3 years ago

Math homework please help new subject dont know it. (trolls = report)

Natasha2012 [34]

Find area on triangle:
A=(1/2)bh
A=(1/2)(8 units)(6 units)
A=(1/2)(48 units^2)
A=24 units^2
Just do the same for the rest of this page following the equation for the area of a triangle A=(1/2)(base)(height)

Find the error:
A=(1/2)bh
100m^2=(1/2)b(20m)
100m^2=(10m)b
10m=b
He forgot to multiply bh by 1/2

Area of parallelogram:
A=bh
A=(3 units)(3 units)
A= 9 units^2
The red line is the height and the bottom of the parallelogram is the base, just count the squares. Imagine cutting the parallelogram at the red line and moving that triangle piece to the other side, it’ll make a rectangle, that’s why the equation is the same.

6 0

3 years ago

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