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Gemiola [76]
3 years ago
9

If x varies inversely as y, and x = 11 when y = 15, find x when y = 3. 11/5 55 33/15

Mathematics
1 answer:
kirill115 [55]3 years ago
4 0

Answer:

Step-by-step explanation:

x=55. Explanation: If x varies inversely as y , then that means: y=kx. Let's multiply both sides of the equation by x to find k . k=xy.

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line J passes through points (-6,1) and (-3,6) . Line m is parallel to line j and passes through point (15,-1).what is the equat
VMariaS [17]
Slope of the line J
(6-1) / (-3-(-6))
5 / 3

so
m = 5/3

If this line passes through (15,-1)
y = mx + c
-1 = (5/3)*15 + c
-1 = 25 + c
c = -26

equation is
y = (5/3)x - 26
7 0
4 years ago
PLEASE HELP ME!! IM IN DESPERATE NEED OF HELP
vesna_86 [32]

Answer:

12

Step-by-step explanation:

6 0
4 years ago
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Luigi has a piece of cloth that is 42 inches long. He wants to divide it into 3 equal pieces. What is the length of each piece?​
Ivanshal [37]

Answer:

if you divide it by 2, you would have 128 slices because 256 ÷ 2 gives you 128, and when you multiply 128 by 2 you get 26.

Step-by-step explanation:

7 0
3 years ago
Which equation represents the polynomial function with zeros −1, 1, and 3 (multiplicity of 2), and a y-intercept of −18? (4 poin
Vlad [161]
Correct Answer:
3rd option is the correct answer

Solution:
The zeros of the polynomial are -1,1 and 3. The multiplicity of 3 is 2. So the polynomial can be expressed as:

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

The y-intercept of the polynomial is -18. This means the polynomial passes through the point (0,-18). Therefore, y must be -18 when x = 0. Using these values of x and y in previous equation we get:

-18=a (0-3)^{2}(0-1)(0+1) \\  \\ 
-18=-9a \\  \\ 
a=2

The final equation of the polynomial becomes:

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

7 0
3 years ago
A car braked with a constant deceleration of 16ft/s2, producing skid marks measuring 200 feet before coming to a stop. How fast
Llana [10]

Answer:the car was traveling at a speed of 80 ft/s when the brakes were first applied.

Step-by-step explanation:

The car braked with a constant deceleration of 16ft/s^2. This is a negative acceleration. Therefore,

a = - 16ft/s^2

While decelerating, the car produced skid marks measuring 200 feet before coming to a stop.

This means that it travelled a distance,

s = 200 feet

We want to determine how fast the car was traveling (in ft/s) when the brakes were first applied. This is the car's initial velocity, u.

Since the car came to a stop, its final velocity, v = 0

Applying Newton's equation of motion,

v^2 = u^2 + 2as

0 = u^2 - 2 × 16 × 200

u^2 = 6400

u = √6400

u = 80 ft/s

7 0
3 years ago
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