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Serhud [2]
3 years ago
5

How do I write 0.35 as money

Mathematics
2 answers:
Rina8888 [55]3 years ago
8 0
You can right it as 35 cents
kondaur [170]3 years ago
8 0
The answer would be 36 cents.
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You will spend 21 dollars each week
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(A)+(B)=8 I need help for this question please, and thanks. :) 10 pts!!
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You haven't included enough information for the question, but assuming that a and b are the same then a is 4 and b is 4
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A rock is thrown upward from a bridge that is 57 feet above a road. The rock reaches its maximum height above the road 0.76 seco
bekas [8.4K]

answer:

f(t) = -9.9788169(t -0.76)^2 +57

Step-by-step explanation:

On this question we see that we are given two points on a certain graph that has a maximum point at 57 feet and in 0.76 seconds after it is thrown, we know can say this point is a turning point of a graph of the rock that is thrown as we are told that the function f determines the rocks height above the road (in feet) in terms of the  number of seconds t since the rock was thrown therefore this turning point coordinate can be written as (0.76, 57) as we are told the height represents y and x is represented by time in seconds. We are further given another point on the graph where the height is now 0 feet on the road then at this point its after 3.15 seconds in which the rock is thrown in therefore this coordinate is (3.15,0).

now we know if a rock is thrown it moves in a shape of a parabola which we see this equation is quadratic. Now we will use the turning point equation for a quadratic equation to get a equation for the height which the format is f(x)= a(x-p)^2 +q  , where (p,q) is the turning point. now we substitute the turning point

f(t) = a(t-0.76)^2 + 57, now we will substitute the other point on the graph or on the function that we found which is (3.15, 0) then solve for a.

0 = a(3.15 - 0.76)^2 + 57

-57 =a(2.39)^2  

-57 = a(5.7121)

-57/5.7121 =a

-9.9788169 = a      then we substitute a to get the quadratic equation therefore f is

f(t) = -9.9788169(t-0.76)^2 +57

4 0
2 years ago
Does 8 + 2(x−6) = −2 + 2x −2 have one solution, no solutions, or infinite solutions.
tia_tia [17]

Answer:

Infinite solutions

Step-by-step explanation:

All solutions are possible.

7 0
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A punch glass is in the shape of a hemisphere with a radius of 5 cm. If the punch is being poured into the glass so that the cha
Galina-37 [17]

Answer:

28.27 cm/s

Step-by-step explanation:

Though Process:

  • The punch glass (call it bowl to have a shape in mind) is in the shape of a hemisphere
  • the radius r=5cm
  • Punch is being poured into the bowl
  • The height at which the punch is increasing in the bowl is \frac{dh}{dt} = 1.5
  • the exposed area is a circle, (since the bowl is a hemisphere)
  • the radius of this circle can be written as 'a'
  • what is being asked is the rate of change of the exposed area when the height h = 2 cm
  • the rate of change of exposed area can be written as \frac{dA}{dt}.
  • since the exposed area is changing with respect to the height of punch. We can use the chain rule: \frac{dA}{dt} = \frac{dA}{dh} . \frac{dh}{dt}
  • and since A = \pi a^2 the chain rule above can simplified to \frac{da}{dt} = \frac{da}{dh} . \frac{dh}{dt} -- we can call this Eq(1)

Solution:

the area of the exposed circle is

A =\pi a^2

the rate of change of this area can be, (using chain rule)

\frac{dA}{dt} = 2 \pi a \frac{da}{dt} we can call this Eq(2)

what we are really concerned about is how a changes as the punch is being poured into the bowl i.e \frac{da}{dh}

So we need another formula: Using the property of hemispheres and pythagoras theorem, we can use:

r = \frac{a^2 + h^2}{2h}

and rearrage the formula so that a is the subject:

a^2 = 2rh - h^2

now we can derivate a with respect to h to get \frac{da}{dh}

2a \frac{da}{dh} = 2r - 2h

simplify

\frac{da}{dh} = \frac{r-h}{a}

we can put this in Eq(1) in place of \frac{da}{dh}

\frac{da}{dt} = \frac{r-h}{a} . \frac{dh}{dt}

and since we know \frac{dh}{dt} = 1.5

\frac{da}{dt} = \frac{(r-h)(1.5)}{a}

and now we use substitute this \frac{da}{dt}. in Eq(2)

\frac{dA}{dt} = 2 \pi a \frac{(r-h)(1.5)}{a}

simplify,

\frac{dA}{dt} = 3 \pi (r-h)

This is the rate of change of area, this is being asked in the quesiton!

Finally, we can put our known values:

r = 5cm

h = 2cm from the question

\frac{dA}{dt} = 3 \pi (5-2)

\frac{dA}{dt} = 9 \pi cm/s// or//\frac{dA}{dt} = 28.27 cm/s

5 0
3 years ago
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