suppose x and y vary inversely and y=10 when x=4. write a function that models the variation then find y when x =90

Mathematics

1 answer:

ankoles [38]3 years ago

4 0

X = k/y
4 = k/10
k = 4 x 10 = 40
Thus, x = 40 / y

90 = 40 / y
y = 40/90 = 4/9

You might be interested in

Find X- Y if X= 1/4 and Y= -X A) 1/3 B) 1/2 C) -1/3 D)-1/2

77julia77 [94]

Answer:

Step-by-step explanation:

x = 1/4

y = -x = -1/4

$x-y=\frac{1}{4}-(-\frac{1}{4})\\\\=\frac{1}{4}+\frac{1}{4}\\\\=\frac{1+1}{4}\\\\=\frac{2}{4}\\\\=\frac{1}{2}$

6 0

3 years ago

The director of a marching band asks the band members to line up in rows of four, but one is left over. Then she tries to line t

Hatshy [7]

There are 81 band members. :-)

4 rows of 20 = 80 ... with one left over = 80 + 1 (81)
6 rows of 13 = 78 ... with three left over = 78 + 3 (81)
7 rows of 11 = 77 ... with four left over = 77 + 4 (81)

5 0

4 years ago

Read 2 more answers

Given the vertices, determine the quadrilaterals most specific classification: Parrellogram, Rectangle, Rhombus, or square.

ankoles [38]

For this case what you must do to solve the problem is to do the following steps:
1) graph each ordered pair.
2) Join each of the ordered pairs.
3) see the type of dormated figure.
Note: see attached image.
Answer:
The quadrilateral is a square.

7 0

3 years ago

Find the volume of the pyramid. Round to the nearest tenth if necessary. hurry!!

Agata [3.3K]

Answer:

Step-by-step explanation:

All pyramids have a volume as one-third of a prism with the same base area and the same height.

This will make the formula for the volume $\frac{1}{3}(base*height)$ . We already know the height of the pyramid is 10, but the area of the base for this rectangle would be its length and its width. Let's calculate this:

$7*2.3=16.1\hspace{0.1cm}in^2$

Let's put these values back into the formula:

$V=\frac{1}{3}(16.1*10)\\ V=\frac{1}{3}(161)\\ V= 53.7\hspace{0.1cm}in^3$

7 0

2 years ago

The length of a rectangle is 5 yd less than twice the width, and the area of the rectangle is 52 yd^2. Find the dimensions of th

Elis [28]

Let the width be w.
The length is 5 yd less than twice the width, so the length is 2w - 5.
The area is LW, and the area is 52 yd^2

LW = A

(2w - 5)(w) = 52

2w^2 - 5w = 52

2w^2 - 5w - 52 = 0

2w^2 - 13w + 8w - 52 = 0

w(2w - 13) + 4(2w - 13) = 0

(2w - 13)(w + 4) = 0

2w - 13 = 0 or w + 4 = 0

2w = 13 or w = -4

w = 6.5 or w = -4

Since a width cannot be a negative number, discard the solution w = -4.

The width is 6.5 yd.
The length is 2w - 5 = 2(6.5) - 5 = 13 - 8 = 8
The length is 8 yd.

The length is 8 yd, and the width is 6.5 yd.

3 0

3 years ago