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Tom [10]
3 years ago
15

Daniel watches 1 hour of television for every 4 hours of homework he does each week. If he did 12 hours of homework last week, h

ow many hours of television did he watch? A. 8 B. 48 C. 12 D. 3
Mathematics
2 answers:
gayaneshka [121]3 years ago
6 0

Answer:

D. 3 h

4 hours of homework is 1 hour of tv, so

12 hours of homework is:

12 : 4 = 3 hours of tv

Gnesinka [82]3 years ago
6 0
The answer is d.3 because 4x3 is 12 and 1x3 is 3
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0.8

Step-by-step explanation:

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A student claims that 8^3 x 8^-5 is greater than 1. Explain whether the student is correct or not.
nlexa [21]
Remember:
a^m * a^n=a^(m+n)
a^-m=1/a^m


8³ x 8⁻⁵=8³⁻⁵=8⁻²=1/8²=1/64<1

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Pls pls help it is due todayyy
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3 years ago
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
The measures of the angles of a triangle are shown in the figure below. Solve for x.
madreJ [45]

Answer:

96x^3+24x^2-432x+270

Step-by-step explanation:

hope it helped

3 0
3 years ago
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