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ValentinkaMS [17]
3 years ago
15

Tell whether the lines for each pair of equations are parallel, perpendicular, or neither. y =-7/2x - 9 -14x - 4y = -20

Mathematics
1 answer:
RideAnS [48]3 years ago
3 0
Parallel. slope is ,-7/2 for both
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When x=12, y=8. Find x when y=12
Digiron [165]
<span>x : y is equal to 12 : 8. These can be reduced or cancelled down to 3 : 2 in the same proportions. Therefore, when y is 12, you have to multiply the smallest value y can be by 6, and must do the same to the smallest x value to retain the proportions. x = 18 when y = 12.</span>
6 0
3 years ago
Fully Factor 3x^2 – 16x + 20
ryzh [129]

Answer:

C. (3x2 - 10) (x - 2)

Step-by-step explanation:

3x^2 - 16x +20

Step 1: Sum = - 16x

Product = 60x^2

Step 2: Find 2 numbers that their sum is -16 and their product is 60. If you do that correctly, then you will get two numbers: 10 and -6

Step 3: Replace -16x with the two numbers found in step 2, then you will have; 3x^2 - 6x - 10x + 20

Step 4: Factorise the equation in step 3, like so;

(3x^2 - 6x)(- 10x + 20)

3x(x - 2) -10(x - 2)

(3x - 10)(x - 2)

To check if the answer is correct, expand the bracket: (3x - 10)(x - 2)

If the bracket is opened properly, you will get 3x^2 - 16x + 20

4 0
3 years ago
1/2 to exponent of 4
Neporo4naja [7]

Answer:

0.0625 is the answer.

Hope it helps you!!!

4 0
3 years ago
Could someone explain how they get the answer for this?
jarptica [38.1K]

Answer:

40°

Step-by-step explanation:

The reference angle is the positive acute angle created by the terminal arm and the x-axis.

The highlighted red in the picture below shows what we're looking for.

The arm rotated 220° (but 'backwards' so the value given is negative).

|-220°| - 2(90°)          <= Subtract two right angles for two quadrants

= 220° - 2(90°)

= 220° - 180°

= 40°

Therefore, the reference angle is 40°.

If you got 50°, you probably calculated the angle with the terminal arm and the y-axis. Remember to always use the nearest side of the x-axis!

4 0
4 years ago
F(x)=x+3 , G(x)=x^2-2<br> (F/G)(-1)-G(3)
trapecia [35]

f(x)=x+3;\ g(x)=x^2-2\\\\\left(\dfrac{f}{g}\right)(-1)-g(3)=?\\\\\left(\dfrac{f}{g}\right)(x)=\dfrac{f(x)}{g(x)}\\\\g(3)=3^2-2=9-2=7\\\\f(-1)=-1+3=2\\\\g(-1)=(-1)^2-2=1-2=-1\\\\\left(\dfrac{f}{g}\right)(-1)-g(3)=\dfrac{2}{-1}-7=-2-7=-9

7 0
3 years ago
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