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oee [108]
3 years ago
9

Mary spent a total of $354.06 for a party. She spent $200 21 on food, plus an additional $30.77 for each hour of the party. How

longwas the party?

Mathematics
1 answer:
BigorU [14]3 years ago
7 0
(354.06 - 200.21) divided by 30.77 = 5
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A group of circus bears made a pyramid. 3 bears fell of the top, leaving 3 on the bottom. How many bears made the pyramid?
kvasek [131]

Answer:

6, right?

Step-by-step explanation:

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3 years ago
At noon, ship A is 130 km west of ship B. Ship A is sailing east at 25 km/h and ship B is sailing north at 20 km/h. How fast is
Hitman42 [59]

Answer:

answer = 12.87 km/h

Step-by-step explanation:

Given

Ship A is sailing east at 25 km/h = \frac{dx}{dt}

ship B is sailing north at 20 km/h =\frac{dy}{dt}

here x and y are the  sailing at t = 4 : 00 pm for ship A and B respectively

so we get x = 4 ×25 =100 km/h

                 y = 4× 20 = 80 km/h

let z is the distance between the ships, we need to find \frac{dz}{dt} at t = 4 hr

At noon, ship A is 130 km west of ship B (12:00 pm)

so equation will be

z^2 = (130-x)^2 + y^2......................(i)\\put x = 100 and y = 80 \\\\we |  | get \\

z^2 = 30^2 + 80^2\\z =\sqrt{7300} km/h

derivative first equation w . r. to t we get

2z\frac{dz}{dt} =-2(130-x)\frac{dx}{dt}+2y\frac{dy}{dt}

\frac{dz}{dt} =\frac{1}{z}[(x -130)\frac{dx}{dt} +y\frac{dy}{dt}]

\frac{dz}{dt} = \frac{( -20\times25 + 80\times20)}{\sqrt{7300} }

     = \frac{1100}{85.44}\\  = 12.87km/h

8 0
3 years ago
Please help <br><br> It’s due tonight
Natalija [7]

Answer: 18 units long

Step-by-step explanation:

In the diagram, it looks like line segment AW is 6 units long and WB is 12 units long. Because of parts-whole postulate (the whole is equal to the sum of its parts), you can add AW and WB's lengths together. 6 + 12 = 18, so the length of line segment AB is 18 units long.

5 0
3 years ago
Sarah has some grapes. She gathers 3 more grapes. She now has 40 grapes.
marysya [2.9K]
Sarah had 37 grapes to begin with

40-3=37
3 0
2 years ago
Geraldine is asked to explain the limits on the range of an exponential equation using the function f(x) = 2^x. She makes these
Dafna11 [192]

Consider the exponential function f(x)=2^x.

By definition, the domain of a function is the set of input argument values for which the function is real and defined.

Let we take x=2, then f(2)=2^2=4

x=4, then f(4)=2^4=16

x=10, then f(10)=2^{10}=1024

If we chose larger values of x, we get larger function values.

For example, If we take f(0)=2^0=1

f(-3)=2^{-3}=\frac{1}{2^3}=\frac{1}{8}

f(-10)=2^{-10}=\frac{1}{2^10}=\frac{1}{1024}

Thus if we choose smaller and smaller values of x. the f unction values will be smaller and smaller functions.

Thus the domain of the function is the set of all real numbers.

Thus the range is limited to the set of positive real numbers. That is, (0,\infty)

If we choose larger values of x, we will get larger function values, as the function values will be larger powers of 2.

If we choose smaller and smaller x values, the function values will be smaller and smaller fractions.

8 0
3 years ago
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