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hichkok12 [17]
3 years ago
8

Can someone help me please.....

Mathematics
1 answer:
Helen [10]3 years ago
5 0

Answer:

the third one

Step-by-step explanation:

2 1/2

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Evaluate the expression for the given value of the variable 1. 8q + 6; q = 2 2. 9 + 4g; g = -3 3. 7x - 7 ; x = 4 4. 10 - 3.2n; n
BlackZzzverrR [31]

Evaluate the expression for the given value of the variable

1. 8q + 6; q = 2

Plug in 2 for q in the given expression

8(2) + 6= 22

2. 9 + 4g; g = -3

9 + 4(-3) = 9 - 12 = -3

3. 7x - 7 ; x = 4

7(4) - 7 = 28 -7 = 21

4. 10 - 3.2n; n = 2

10 - 3.2(2) = 10 - 6.4 = 3.6

5. 12p + 4; p = 1.5

12(1.5) + 4= 18  + 4 = 22

6. 5w + 10 = 40

Subtract both sides by 10

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w = 6

7. 4y - 12 = 60

Add both sides by 12

4y = 72

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8. 10k + 5 = 65

Subtract both sides by 5

10k = 60, now divide both sides by 10

k = 6

9. 2b - 15 = 33

Add both sides by 15

2b = 48

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6 0
3 years ago
If x^2 + y^2 = 25, find the equation of the line tangent to the circle when x = -4
bekas [8.4K]

Answer:

In the point (-4,3) tangent is : y=4/3x+25/3 i.e, y=1.33x+8.33

In the point (-4,-3) tangent is: y=-4/3x-25/3 i.e, y=-1.33x-8.33

Step-by-step explanation:

The center of the circle is in (0,0), because x^2+y^2=(x-0)^2+(y-0)^2=25. And the radius is 5, because 5^2=25.

We have first center: (x1,y1)=(0,0)

The point of the tangent is in x=-4, so x^2+y^2=25 is 16+y^2=25,i.e y=3 and y=-3. Like you can see we have 2 tangents to the circle when x=-4. the point (-4,3) and (-4,-3).

We will do for the first point:

The point of tangent is (x2,y2)=(-4,3)

The gradient  of the radius

mr=(y2-y1)/(x2-x1)=(3-0)/(-4-0)=-3/4

The gradient of the tangent mt is:

mr*mt=-1 (they are ortogon)

mt=-1/mr=-1/(-3/4)=4/3

the equation of the line trought one point is:

y-y2=k(x-x2), where k is mt,i.e. k=4/3

y-3=4/3(x-(-4))

y=4/3x+16/3+3

y=4/3x+25/3

y=1.33x+8.33

Now for the 2nd point (-4,-3),

mr=-3/-4=3/4

mt=-4/3

y+3=-4/3(x-(-4))

y=-4/3x-16/3-3

y=-4/3x-25/3

y=-1.33x-8.33

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Step-by-step explanation:

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