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valkas [14]
3 years ago
12

The water level in a lake was 22 inches below normal at the beginning of March. The water level decreased by 3 1/6 inches in Mar

ch and increased by 1 6/7 inches in April
Mathematics
1 answer:
Vinvika [58]3 years ago
8 0
At the beginning it was 22 inches below normal (-22).
Then it decreased by 3 1/6 (-19/6).
Then increased by 1 6/7 (+13/7)

-22 - 19/6 + 13/7
-924/42 - 133/42 + 78/42
-1057/42 + 78/42
-979/42
-23 13/42

It was 23 13/42 inches below normal after April.
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liraira [26]

Answer:

saving money

Step-by-step explanation:

it wouldn't be none with repaid yet!

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2 years ago
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What formula can we use to find the circumference of a circle?
Assoli18 [71]

Answer:

B plz mrk brainliest

Step-by-step explanation:

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3 years ago
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With a family bowling pass, families can bowl for $4 per game . The pass cost $10 per year. Use equation , a table and a graph t
Klio2033 [76]

Answer:

Part a) a=4g+10 (see the explanation)

Part b) see the explanation

Part c) The graph in the attached figure

Step-by-step explanation:

<u><em>The complete question is</em></u>

With a family bowling pass, families can bowl for $4 per game. The pass costs $10 per year. Use an equation, a table, and a graph to explain the relationship between the total amount of money spent on bowling in a year, a, and the number of games a family plays in a year, g. Part A Use words and an equation to represent this problem. Part B Create a table to show values for g and a. Part C Use the values from your table to draw a graph.

Part A) Use words and an equation to represent this problem

Let

a ----> the total amount of money spent on bowling in a year

g ---> the number of games a family plays in a year

we know that

The linear equation in slope intercept form is equal to

a=mg+b

where

m is the slope or unit rate

b is the a-intercept or initial value

In this problem we have

m=\$4\ per\ game

b=\$10

substitute

a=4g+10

Part b) Create a table to show values for g and a

Assume different values of g and calculate the corresponding values of a

For g=0 ----> a=4(0)+10=\$10

For g=1 ----> a=4(1)+10=\$14

For g=2 ----> a=4(2)+10=\$18

For g=3 ----> a=4(3)+10=\$22

For g=4 ----> a=4(4)+10=\$26

The table is

\left[\begin{array}{ccc}g&a\\0&10\\1&14\\2&18\\3&22\\4&26\end{array}\right]

Part c) Use the values from your table to draw a graph

we have the ordered pairs

(0,10),(1,14),(2,18),(3,22),(4,26)

Plot the ordered pairs and join them to graph the line

The graph in the attached figure

6 0
3 years ago
A parabola can be drawn given a focus of
Volgvan

Answer:

y=-\frac{1}{4}(x-2)^{2}+9

Step-by-step explanation:

Any point on a given parabola is equidistant from focus and directrix.

Given:

Focus of the parabola is at (2,8).

Directrix of the parabola is y=10.

Let (x,y) be any point on the parabola. Then, from the definition of a parabola,

Distance of (x,y) from focus = Distance of (x,y) from directrix.

Therefore,

\sqrt{(x-2)^{2}+(y-8)^{2}}=|y-10|

Squaring both sides, we get

(x-2)^{2}+(y-8)^{2}=(y-10)^{2}\\(x-2)^{2}=(y-10)^{2}-(y-8)^{2}\\(x-2)^{2}=(y-10+y-8)(y-10-(y-8))...............[\because a^{2}-b^{2}=(a+b)(a-b)]\\(x-2)^{2}=(2y-18)(y-10-y+8)\\(x-2)^{2}=2(y-9)(-2)\\(x-2)^{2}=-4(y-9)\\y-9=-\frac{1}{4}(x-2)^{2}\\y=-\frac{1}{4}(x-2)^{2}+9

Hence, the equation of the parabola is y=-\frac{1}{4}(x-2)^{2}+9.

4 0
3 years ago
INFO: The chairs are evenly spaced around
lana66690 [7]

Answer: About 72 feet

The more accurate value is roughly 71.7260397 feet (however, this value isn't fully exact either).

======================================================

Explanation:

Check out the diagram below. It's effectively the same diagram your teacher provided, but I've added points M and N. I've left out any unnecessary lines.

  • M = center of the circle
  • N = location where the cable is anchored to the ground

We're told that "(the length of cable from) E to the ground is 35 feet", so this is the red portion of cable that I've marked this as NE = 35, ie segment NE is 35 feet long. We're also told that NF = 40 feet, which is also shown in red.

The blue portions are chords of the circle. We're given one chord as LE = 80 ft, while the other chord JF is what we want to find out.

-----------------

What we need to find is the radius of this circle. Right away, it's not clear what the radius is; however, we can use the chord LE = 80 to help find it.

It turns out that the chord length c is connected to the radius r and central angle theta like so:

c = 2*r*sin(0.5*theta)

where theta must be in radian mode. The central angle subtends the arc that forms the chord in question.

If we started at point L, and counted the number of spaces to get to E while going clockwise, then we can see that there are 5 spaces

Those five spaces could be written as

  1. The jump from L to A
  2. The jump from A to B
  3. The jump from B to C
  4. The jump from C to D
  5. The jump from D to E

Or you could mark it as such on the diagram for a visual reference. Overall, the circle has been cut into 12 equal slices. So going from L to E, going clockwise, will have us take up 5/12 of the full circle.

There are 2pi radians in a full circle, meaning the central angle LME is (5/12)*2pi = 5pi/6 radians.

-----------------

Plug theta = 5pi/6 and c = 80 into the formula mentioned. Isolate the radius.

c = 2*r*sin(0.5*theta)

80 = 2*r*sin(0.5*5pi/6)

80 = 2*r*sin(5pi/12)

80 = 2*r*0.9659258

80 = 2*0.9659258*r

80 = 1.9318516r

1.9318516r = 80

r = 80/1.931 8516

r = 41.4110483

The radius is approximate. The radius is in feet. Make sure your calculator is in radian mode.

-----------------

Now that we know the radius, we can determine how long chord JF is.

The central angle for chord JF is angle JMF

If we start at J and go to F, along the shorter path, then we've gone 4 spaces. This is 4/12 = 1/3 of the full circle. So the radian measure of angle JMF is (1/3)*2pi = 2pi/3 radians.

So,

c = 2*r*sin(0.5*theta)

c = 2*41.4110483*sin(0.5*2pi/3)

c = 71.7260397

Chord JF is roughly 71.7260397 feet long.

When rounding to the nearest foot, that's about 72 feet.

6 0
2 years ago
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