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slavikrds [6]
3 years ago
6

Zain bought ice cream and sherbet at the grocery store and is placing the containers in the freezer. The container of ice cream

was sold in a 10" × 14" × 12" container, and the sherbet was sold in an 11" × 18" × 9" container. How much less space does the container of ice cream occupy than the container of sherbet? *
Mathematics
1 answer:
lianna [129]3 years ago
7 0

Answer:

102 in^3.

Step-by-step explanation:

Volume of the  ice cream container = 10*14*12 = 1680 in^3.

Volume of the  sherbet container = 11*18^9 = 1782 in^3.

So the answer is 1782 - 1680

= 102 in^3.

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6 quarters and 6 dimes

Step-by-step explanation:

since there is 6 quarters it would be 6x$.25 which is $1.50, and each dime is worth 10 cents so 6x$.10 is $.60, and $1.50+$.60=$2.10

5 0
2 years ago
Find all points on the curve x=4cos(t),y=4sin(t) that have the slope of 12.
Marianna [84]

Answer:

\left (-\dfrac{4}{\sqrt{5}},\dfrac{8}{\sqrt{5}}\right )\text{ and }\left (\dfrac{4}{\sqrt{5}},-\dfrac{8}{\sqrt{5}}\right ).

Step-by-step explanation:

We need to find all the points on the curve x=4cos(t),y=4sin(t) that have the slope of 1/2.

x=4cos (t)

\dfrac{dx}{dt}=-4sin (t)

y=4sin (t)

\dfrac{dy}{dt}=4cos (t)

Now,

\dfrac{dy}{dx}=\dfrac{dy}{dt}\times \dfrac{dt}{dx}

\dfrac{dy}{dx}=4cos (t)\times \dfrac{1}{-4sin (t)}

\dfrac{dy}{dx}=-\cot t

So, slope of the curve is -\cot t.

-\cot t=\dfrac{1}{2}

-\tan t=2

\tan t=-2            ...(1)

Using \sec^2t=1+\tan^2t, we get

\sec^2t=1+(-2)^2

\sec^2t=1+4

\sec t=\pm \sqrt{5}

\cos t=\pm \dfrac{1}{\sqrt{5}}

Now,

\sin^2t=1-cos^2t

\sin t=\pm \sqrt{1-\dfrac{1}{5}}

\sin t=\pm \sqrt{\dfrac{4}{5}}

\sin t=\pm \dfrac{2}{\sqrt{5}}

It equation (1), tan(t) is negative. So, sin and cos have different signs.

If \sin t= \dfrac{2}{\sqrt{5}}, then \cos t=- \dfrac{1}{\sqrt{5}}.

x=4cos (t)=-\dfrac{4}{\sqrt{5}}

y=4sin (t)=\dfrac{8}{\sqrt{5}}

If \sin t=- \dfrac{2}{\sqrt{5}}, then \cos t= \dfrac{1}{\sqrt{5}}.

x=4cos (t)=\dfrac{4}{\sqrt{5}}

y=4sin (t)=-\dfrac{8}{\sqrt{5}}

Therefore, the two points are \left (-\dfrac{4}{\sqrt{5}},\dfrac{8}{\sqrt{5}}\right )\text{ and }\left (\dfrac{4}{\sqrt{5}},-\dfrac{8}{\sqrt{5}}\right ).

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Answer:

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