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77julia77 [94]
3 years ago
9

You want to put a 5 inch thick layer of topsoil for a new 16 ft by 34 ft garden. The dirt store sells by the cubic yards. How ma

ny cubic yards will you need to order? The store only sells in increments of 1/4 cubic yards.
Mathematics
1 answer:
shepuryov [24]3 years ago
5 0

Answer: About 8.5 cubic yards

Step-by-step explanation:

Given : The length of the garden = 16 ft.

The width of the garden = 34 ft.

The depth of the thick layer of topsoil on the garden = 5 inch

=\dfrac{5}{12}\text{ ft.}               [Since 1 foot = 12 inches]

The volume of a rectangular prism :-

V=l*w*h, where l is length , w is width and h is height.

The number of cubic feet of topsoil required will be

V=16\times34\times\dfrac{5}{12}=\dfrac{680}{3}\text{cubic feet}

Since 1 yard = 3 feet

1\text{ foot}=\dfrac{1}{3}\text{ yard}

V=\dfrac{680}{3}\times\dfrac{1}{3}\times\dfrac{1}{3}\times\dfrac{1}{3}=8.3950617284\approx8.50\text{cubic yards}

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Three mulitplied by five-hundred eighty four

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For the second week of June, Kevin Adams worked 46 hours. Kevin earns $18.40 an hour. His employer pays overtime for all hours w
zimovet [89]

<u>Answer:</u>

$165.60 overtime money only

$736 regular time only

added together (gross income)

$901.60

<u>Step-by-step explanation:</u>

  • 46hrs
  • 18.40 an hr
  • hrs <em>over</em> 40 = overtime= *1.5

To figure out if there is any overtime here, we <em>subtract</em>.

46-40= 6hrs of overtime

To figure out the amount of cash gained for regular time, we <em>multiply.</em>

40*18.40 = 736

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

We need to know how much money is paid per overtime hour.

18.40*1.5 = 27.60

To figure out the amount of cash gained during overtime, we <em>multiply.</em>

6*27.60 = 165.60

<em>add</em> for gross income (total earnings)

165.60+736 = 901.60

7 0
2 years ago
What are the properties used to solve 4(x+3)=20
ryzh [129]

Answer:

Multiproperty

Step-by-step explanation:

To solve this problem, we use different properties to arrive at our answer.

  Given expression;

               4(x + 3) = 20;

   First we use the distributive property to expand;

              4x + 12  = 20

       Now use subtraction property by adding (-12) to both sides;

              4x + 12 + (-12)  = 20 + (-12)

              4x  = 8

           you can now divide by 4,

                   x  = 2

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2 years ago
How many solutions does the equation have?
Studentka2010 [4]
The answer is c no solution
6 0
3 years ago
In right ABC, AN is the altitude to the hypotenuse. FindBN, AN, and AC,if AB =2 5 in, and NC= 1 in.
Rama09 [41]

From the statement of the problem, we have:

• a right triangle △ABC,

,

• the altitude to the hypotenuse is denoted AN,

,

• AB = 2√5 in,

,

• NC = 1 in.

Using the data above, we draw the following diagram:

We must compute BN, AN and AC.

To solve this problem, we will use Pitagoras Theorem, which states that:

h^2=a^2+b^2\text{.}

Where h is the hypotenuse, a and b the sides of a right triangle.

(I) From the picture, we see that we have two sub right triangles:

1) △ANC with sides:

• h = AC,

,

• a = ,NC = 1,,

,

• b = NA.

2) △ANB with sides:

• h = ,AB = 2√5,,

,

• a = BN,

,

• b = NA,

Replacing the data of the triangles in Pitagoras, Theorem, we get the following equations:

\begin{cases}AC^2=1^2+NA^2, \\ (2\sqrt[]{5})^2=BN^2+NA^2\text{.}\end{cases}\Rightarrow\begin{cases}NA^2=AC^2-1, \\ NA^2=20-BN^2\text{.}\end{cases}

Equalling the last two equations, we have:

\begin{gathered} AC^2-1=20-BN^2.^{} \\ AC^2=21-BN^2\text{.} \end{gathered}

(II) To find the values of AC and BN we need another equation. We find that equation applying the Pigatoras Theorem to the sides of the bigger right triangle:

3) △ABC has sides:

• h = BC = ,BN + 1,,

,

• a = AC,

,

• b = ,AB = 2√5,,

Replacing these data in Pitagoras Theorem, we have:

\begin{gathered} \mleft(BN+1\mright)^2=(2\sqrt[]{5})^2+AC^2 \\ (BN+1)^2=20+AC^2, \\ AC^2=(BN+1)^2-20. \end{gathered}

Equalling the last equation to the one from (I), we have:

\begin{gathered} 21-BN^2=(BN+1)^2-20, \\ 21-BN^2=BN^2+2BN+1-20 \\ 2BN^2+2BN-40=0, \\ BN^2+BN-20=0. \end{gathered}

(III) Solving for BN the last quadratic equation, we get two values:

\begin{gathered} BN=4, \\ BN=-5. \end{gathered}

Because BN is a length, we must discard the negative value. So we have:

BN=4.

Replacing this value in the equation for AC, we get:

\begin{gathered} AC^2=21-4^2, \\ AC^2=5, \\ AC=\sqrt[]{5}. \end{gathered}

Finally, replacing the value of AC in the equation of NA, we get:

\begin{gathered} NA^2=(\sqrt[]{5})^2-1, \\ NA^2=5-1, \\ NA=\sqrt[]{4}, \\ AN=NA=2. \end{gathered}

Answers

The lengths of the sides are:

• BN = 4 in,

,

• AN = 2 in,

,

• AC = √5 in.

7 0
1 year ago
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