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3 years ago

12

Hannah decided to make finger gelatin for a huge children’s party. She had to make 8 packs of gelatin. Each pack needed 2 cups o

f water. How many quarts of water will she need?
8 quarts
4 quarts
6 quarts
3 quarts

2 answers:

hjlf3 years ago

8 0

The answer is 4 quarts

Nadya [2.5K]3 years ago

4 0

Answer:

4 quarts

Step-by-step explanation:

need 16 cups. In quarts that would be 4 quarts.

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The outside diameter of the igloo is 12 feet. The thickness of each block of ice that was used to construct the igloo is 1.5 fee

mojhsa [17]

The first thing we are going to do for this case is to model the igloo as a hemisphere.
The volume of the hemisphere is:
V = ((4/3) * (pi) * (r ^ 3)) / 2
Where,
r: sphere radio
We have then that the internal diameter is:
di = 12-1.5
di = 10.5 feet
So the radio is:
r = di / 2
r = 10.5 / 2
r = 5.25 feet
Substituting values:
V = ((4/3) * (pi) * ((5.25) ^ 3)) / 2
V = 303.1 feet ^ 3
Answer:
the amount of space of the living area inside the igloo is:
V = 303.1 feet ^ 3

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3 years ago

kifflom [539]

20 cats * ( 1 bag of food / 5 cats per month ) = 4 bags of food per month to feed all 20 cats

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3 years ago

Read 2 more answers

Use the multiplication property of exponents to determine if each statement is true or false.

lisabon 2012 [21]

1)
LHS:
${a}^{4} \times {a}^{3} \times a \\ = {a}^{4 + 3 + 1} \\ = {a}^{7}$
LHS = RHS. So, the statement is true

2)
LHS:
${5}^{2} \times {3}^{4} \\ = 25 \times 81 \\ = {5}^{2} \times {9}^{2} \\ = {45}^{2}$
LHS not equal to RHS. So, the statement is false

3)
LHS:
${x}^{6} \times {x}^{0} \times {x}^{2} \times {x}^{3} \\ = {x}^{6 + 0 + 2 + 3} \\ = {x}^{11}$
LHS = RHS. So, the statement is true.

4)
LHS:
$7 {m}^{2} \times 2m \times 4 {m}^{5} \\ = (7 \times 2 \times 4) {m}^{2 + 1 + 5} \\ = 56 {m}^{8}$
LHS = RHS. So, the statement is true.

7 0

3 years ago

What is the solution to log (9x)-log 3 = 3?

xeze [42]

Answer:

The solution x = $\frac{e^{3} }{3}$

Step-by-step explanation:

<u>Step 1</u>:-

Given $log(9 x) -log 3 =3$

we have to use formula $log a-log b=log\frac{a}{b}$

$log(9 x)- log 3 = log(\frac{9 x}{3} )$ = $3$

Again by using formula $log a =b$

$a = e^{b}$

so $\frac{9 x}{3} =e^{3}$

<u>step 2:</u>-

$3 x = e^{3}$

$x = \frac{e^{3} }{3}$

The solution of given problem is $x=\frac{e^3}{3}$

5 0

3 years ago

How does the graph of f(x)=3lx+2l+4 relate to its parent function?

Sergeeva-Olga [200]

$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\

\begin{array}{rllll} 
% left side templates
f(x)=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
y=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
f(x)=&{{ A}}\sqrt{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}}(\mathbb{R})^{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}} sin\left({{ B }}x+{{ C}} \right)+{{ D}}
\end{array}$

$\bf \begin{array}{llll}
% right side info
\bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\\\
\bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}
\\\\
\bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\
\end{array}$

$\bf \begin{array}{llll}


\bullet \textit{ vertical shift by }{{ D}}\\
\qquad if\ {{ D}}\textit{ is negative, downwards}\\\\
\qquad if\ {{ D}}\textit{ is positive, upwards}\\\\
\bullet \textit{ period of }\frac{2\pi }{{{ B}}}
\end{array}$

now, with that template above in mind, let's see this one

$\bf parent\implies f(x)=|x|
\\\\\\
\begin{array}{lllcclll}
f(x)=&3|&1x&+2|&+4\\
&\uparrow &\uparrow &\uparrow &\uparrow \\
&A&B&C&D
\end{array}$

A=3, B=1, shrunk by AB or 3 units, about 1/3
C=2, horizontal shift by C/B or 2/1 or just 2, to the left
D=4, vertical shift upwards of 4 units

check the picture below

7 0

3 years ago