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2 years ago

6

The cake that Larry is baking requires 1 cup of sugar. Larry is adding sugar to a bowl using a measuring scoop. In the first sco

op, he adds 7/8 cup of sugar. In the second scoop, he adds another 7/8 cup of sugar. Has Larry measured the sugar correctly?

1 answer:

sasho [114]2 years ago

4 0

No. He added 14/8 cups, or 1 and 3/4th cup.

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A right circular cylinder is inscribed in a sphere with diameter 4cm as shown. If the cylinder is open at both ends, find the la

SOVA2 [1]

Answer:

$8\pi\text{ square cm}$

Step-by-step explanation:

Since, we know that,

The surface area of a cylinder having both ends in both sides,

$S=2\pi rh$

Where,

r = radius,

h = height,

Given,

Diameter of the sphere = 4 cm,

So, by using Pythagoras theorem,

$4^2 = (2r)^2 + h^2$ ( see in the below diagram ),

$16 = 4r^2 + h^2$

$16 - 4r^2 = h^2$

$\implies h=\sqrt{16-4r^2}$

Thus, the surface area of the cylinder,

$S=2\pi r(\sqrt{16-4r^2})$

Differentiating with respect to r,

$\frac{dS}{dr}=2\pi(r\times \frac{1}{2\sqrt{16-4r^2}}\times -8r + \sqrt{16-4r^2})$

$=2\pi(\frac{-4r^2+16-4r^2}{\sqrt{16-4r^2}})$

$=2\pi(\frac{-8r^2+16}{\sqrt{16-4r^2}})$

Again differentiating with respect to r,

$\frac{d^2S}{dt^2}=2\pi(\frac{\sqrt{16-4r^2}\times -16r + (-8r^2+16)\times \frac{1}{2\sqrt{16-4r^2}}\times -8r}{16-4r^2})$

For maximum or minimum,

$\frac{dS}{dt}=0$

$2\pi(\frac{-8r^2+16}{\sqrt{16-4r^2}})=0$

$-8r^2 + 16 = 0$

$8r^2 = 16$

$r^2 = 2$

$\implies r = \sqrt{2}$

Since, for r = √2,

$\frac{d^2S}{dt^2}=negative$

Hence, the surface area is maximum if r = √2,

And, maximum surface area,

$S = 2\pi (\sqrt{2})(\sqrt{16-8})$

$=2\pi (\sqrt{2})(\sqrt{8})$

$=2\pi \sqrt{16}$

$=8\pi\text{ square cm}$

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3 years ago

A square poster has a side length of 26 in. Drawn on the poster are four identical triangles. Each triangle has a base of 8 in.

Firdavs [7]

To find the probability of landing on a triangle, you will want find the combined areas of the triangles and the total area of the square target.

Divide the area of the combined areas and the total area to find the probability of landing on a triangle.

A = 1/2bh
1/2 x 8 x 8
A = 32 square inches
32 x 4
128 square inches (areas of triangles)

A = bh
26 x 26
A = 676 square inches

128/676 = 0.189

There is an approximate probability of 0.19 of hitting a triangle.

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The answer to the question is 5

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Answer:

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