All categories

3 years ago

Three boxes are stacked one on top of the other. One box is

1 answer:

6 0

First, convert all measurements to inches.

Box 1:
3ft(12in)=36in+6in=42in

Box 2:
4ft(12in)=48in+10in=58in

Box 3:
4ft(12in)=48in+5in=53in

Now, add the heights together.

42in+58in+53in=153in

Finally, convert added height to feet.

153in/12in=12 with a remainder of 9, so your final answer is 12ft 9in

You might be interested in

Ms. Nina has 10 lbs of 25% sugar syrup. How much water does she need to add to make 10% sugar syrup?

ryzh [129]

She has 10lbs of 25% syrup... so, in the 10lbs, 25% of that is syrup, the rest, namely the 75% remaining is water or other substances.

let's say she adds "x" lbs of water, to get "y" lbs for the 10% mixture.

how much is 25% of 10lbs? well, (25/100) * 10, or 2.5.

the water has no sugar syrup in it, so is just pure water, so the amoun of syrup in it is 0%, how much is 0% of "x" lbs? well, (0/100) * x, or 0.00x, which is just 0.

how much is 10% of "y" lbs? well (10/100) * y, or 0.10y.

whatever "x" and "y" are, we know that 10 + x = y.

we also know that the syrup amount in that is also 2.5 + 0.00x = 0.10y, thus

$\bf \begin{array}{lccclll}
&\stackrel{lbs}{syrup}&\stackrel{concentration~\%}{syrup}&\stackrel{concentration}{amount}\\
&------&------&------\\
\textit{25\% syrup}&10&0.25&2.5\\
\textit{pur water}&x&0.00&0.00x\\
------&------&------&------\\
\textit{10\% mixture}&y&0.10&0.10y
\end{array}
\\\\\\
\begin{cases}
10+x=\boxed{y}\\
2.5+0.00x=0.10y\\
----------\\
2.5 = 0.10\left( \boxed{10+x} \right)
\end{cases}
\\\\\\
2.5=1 + 0.10x\implies 1.5=0.10x
\\\\\\
\cfrac{1.5}{0.10}=x\implies 15=x$

8 0

3 years ago

Read 2 more answers

A copy machine makes 84 copies in 3 minutes and 30 seconds, how many copies does it make per minute.

matrenka [14]

Answer:

Step-by-step explanation:

3 minutes and 30 seconds = 3.5 minutes

84/3.5 = 24copies/minute

4 0

2 years ago

Read 2 more answers

What is the answer to 118÷12 ?

svetlana [45]

The anwser is 9.8333333333

3 0

3 years ago

Read 2 more answers

50 POINTS

Tcecarenko [31]

Answer:

Part 1: C. $3,159.30

Part 2. C. –5; –135; –10,935

Step-by-step explanation:

Part 1:

Price of the boat = $ 16,600

Depreciation rate = 14% = 0.14

Time of utilization of the boat = 11 years

Price of the boat after 11 years = Original price * (1 - Depreciation rate)^Time of utilization of the boat

Price of the boat after 11 years = 16,600 * (1 - 0.14)¹¹

Price of the boat after 11 years = 16,600 * 0.1903

Price of the boat after 11 years = $ 3,159.30

Part 2:

Let's find out the first term of the sequence given:

A(1) = -5 * 3¹⁻¹

A(1) = -5 * 1

A(1) = -5

Let's find out the fourth term of the sequence given:

A(4) = -5 * 3⁴⁻¹

A(4) = -5 * 3³

A(4) = -5 * 27

A(4) = -135

Let's find out the eighth term of the sequence given:

A(8) = -5 * 3⁸⁻¹

A(8) = -5 * 3⁷

A(8) = -5 * 2,187

A(8) = -10,935

4 0

2 years ago

Y=-x^2+2x+10 y=x+2 Substitution Please show your work Need ASAP

erik [133]

Answer:

The solutions of the system of equations are the points

$(\frac{1-\sqrt{33}} {2},\frac{5-\sqrt{33}} {2})$

$(\frac{1+\sqrt{33}} {2},\frac{5+\sqrt{33}} {2})$

Step-by-step explanation:

we have

$y=-x^{2} +2x+10$ ----> equation A

$y=x+2$ ----> equation B

Solve the system by substitution

substitute equation B in equation A

$x+2=-x^{2} +2x+10$

solve for x

$-x^{2} +2x+10-x-2=0$

$-x^{2} +x+8=0$

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$-x^{2} +x+8=0$

$a=-1\\b=1\\c=8$

substitute in the formula

$x=\frac{-1\pm\sqrt{1^{2}-4(-1)(8)}} {2(-1)}$

$x=\frac{-1\pm\sqrt{33}} {-2}$

$x=\frac{-1+\sqrt{33}} {-2}$ -----> $x=\frac{1-\sqrt{33}} {2}$

$x=\frac{-1-\sqrt{33}} {-2}$ -----> $x=\frac{1+\sqrt{33}} {2}$

Find the values of y

For $x=\frac{1-\sqrt{33}} {2}$

$y=x+2$

$y=\frac{1-\sqrt{33}} {2}+2$ ----> $y=\frac{5-\sqrt{33}} {2}$

For $x=\frac{1+\sqrt{33}} {2}$

$y=x+2$

$y=\frac{1+\sqrt{33}} {2}+2$ ----> $y=\frac{5+\sqrt{33}} {2}$

therefore

The solutions of the system of equations are the points

$(\frac{1-\sqrt{33}} {2},\frac{5-\sqrt{33}} {2})$

$(\frac{1+\sqrt{33}} {2},\frac{5+\sqrt{33}} {2})$

5 0

2 years ago