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mixer [17]
2 years ago
7

If the diameter of a circle is 19ft what is the circumference?

Mathematics
1 answer:
Digiron [165]2 years ago
7 0
The circumference will be 59.69
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Am I right ? Because I need help with this please thanks.
bazaltina [42]

to find the hypotenuse (or the ramp in question 4) you can use the equation

a^{2} +b^{2} =c^{2}

with a and b being the 3 feet and 9 feet and c being the unknown (ramp).

So if you plug 3 and 9 into the equation, it looks like this.

3^{2} +9^{2} =c^{2}

then square them.

9+81=c^{2}

simplify.

90=c^{2}

take the square root of both sides.

\sqrt{90} = c

\sqrt{90} is approx. 9.5 so the answer would be H.

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3 years ago
PLSS HELP IMMEDIATELY!!!! ILL MARK BRAINIEST IF U DONT LEAVE A LINK OR GUESS!!!
stiv31 [10]

Answer:

20 cubic inches

Step-by-step explanation:

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Leona [35]

Answer:

the first blank is 2, the second one is 1

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2 years ago
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Graph the line with slope -1 passing through the point (-1,-4)
Lyrx [107]

Answer: look at pic

Step-by-step explanation:

y=mx+b

m:slope

b:y-intercept

m=-1 (slope) given in the question

y=-1x+b

Plug in the given point (-1,-4) then solve for b

-4=-1(-1)+b

b=-5

y=-x-5

then using the x-intercept and y-intercept you could graph the line

y-intercept = b = -5 (0,-5)

x-intercept (plug in y=0)

0=-x-5

x=-5 (-5,0)

4 0
3 years ago
Suppose a cold beer at 40°F is placed into a warm room at 70°F. Suppose 10 minutes later, the temperature of the beer is 48°F. U
Yuri [45]

Answer:

69.92°F

Step-by-step explanation:

Given:

Initial temperature ( i.e at time, t = 0) = 40°F

Temperature of the room = 70°F

Temperature after 10 minutes ( i.e at time t = 10 ) = 48°F

Now, from Newton's law of cooling

T'(t) = k(A - T(t))

T(t) temperature after time t

T'(t) = \frac{\textup{dT}}{\textup{dt}}

here, A is the room temperature

thus,

\frac{\textup{dT}}{\textup{dt}}  = k(70 - T)

or

\frac{\textup{dT}}{\textup{70-T}} = kdt

on solving the differential equation, we get

T = 70-C^{-kt} ............(1)

Now from the boundary conditions,

 i.e at time, t = 0; T = 40°F

we get,

40 = 70-C^{-k\times0}

or

C = 30

and,

at time, t = 10; T = 48°F

thus,

48 = 70-30^{-k\times10}

or

k = \frac{\textup{-1}}{\textup{10}}ln\frac{11}{15}

or

k = 0.03

Therefore,

for t = 25

from 1 we have

T = 70-30^{-0.03\times25}

or

T = 70 - 0.0780

or

T = 69.92°F

3 0
3 years ago
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