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3 years ago

Hi hhsjsksjsjjsjskakakjdndjsjdnd

1 answer:

8 0

Um hello, you probably shouldn’t be “texting” here because this forum is used only for students who need homework help so um ya

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Please Answer!!!!

ExtremeBDS [4]

First mug holds the most

Solution:

Given that,

You are choosing between two mugs

The volume of cylinder is given as:

$V = \pi r^2h$

Where,

r is the radius and h is the height

One has a base that is 5.5 inches in diameter and a height of 3 inches

$radius = \frac{diameter}{2}$

Therefore,

$r = \frac{5.5}{2}\\\\r = 2.75$

Also, h = 3 inches

Thus volume of cylinder is given as:

$V = \pi \times 2.75^2 \times 3\\\\V = 3.14 \times 22.6875\\\\V = 71.23875 \approx 71.24$

Thus first mug holds 71.24 cubic inches

The other has a base of 4.5 inches in diameter and a height of 4 inches

$Radius = \frac{4.5}{2}\\\\r = 2.25$

h = 4 inches

Therefore,

$V = 3.14 \times 2.25^2 \times 4\\\\V = 3.14 \times 20.25\\\\V = 63.585$

Thus the second mug holds 63.585 cubic inches

On comparing, volume of both mugs,

Volume of first mug > volume of second mug

First mug holds the most

8 0

4 years ago

Write whether each statement is true of false. Explain your reasoning.

sp2606 [1]

6 - p....p = 1.5
6 - 1.5 = 4.5
TRUE...because when u sub in 1.5 for p, the difference is 4.5, which is less then 5.

(30)(420) = 12000 - 600
12600 = 11400...incorrect
FALSE...because the product of 30 and 420 does not equal 600 less then 12000

16.3 + 11.9 < 27
28.2 < 27..incorrect
FALSE...because the sum of 16.3 and 11.9 is greater then 27

5 0

3 years ago

PLEASE HELP

Mumz [18]

If you use/make a number line, then you will see that the answer is -18 or a. When using a number line, negative = ← and positive = →.

6 0

3 years ago

Solve for y x=y^2+4y

Lorico [155]

$x=y^2+4y\\ y^2+4y-x=0\\\Delta=4^2-4\cdot1\cdot(-x)=16+4x\\\\
1.\ \Delta0\\
\sqrt{\Delta}=\sqrt{16+4x}=\sqrt{4(4+x)}=2\sqrt{x+4}\\
y_1=\frac{-4-2\sqrt{x+4}}{2\cdot1}=-2-\sqrt{x+4}\\
y_2=\frac{-4+2\sqrt{x+4}}{2\cdot1}=-2+\sqrt{x+4}\\$

-----------------------------------------------------

$x=y^2+4y\\ y^2+4y-x=0\\
y^2+4y+4-4-x=0\\
(y+2)^2=x+4\\
y+2=\sqrt{x+4} \vee y+2=-\sqrt{x+4}\\
y=-2+\sqrt{x+4} \vee y=-2-\sqrt{x+4}\\$

4 0

4 years ago

Solve the following quadratic-linear system of equations.

Reika [66]

The answer should and must be Done

7 0

3 years ago