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stiks02 [169]
3 years ago
12

Lee washes the outsides of houses. It takes him 40 minutes to power wash a one-story home, and he uses 180 gallons of water. Pow

er washing a two-story home takes 90 minutes and uses 300 gallons of water. Lee works no more than 40 hours each week, and he uses a maximum of 5,000 gallons of water per week. He charges $90 to wash a one-story home and $150 to wash a two-story home. Lee wants to maximize his weekly washing income washing houses. Let x represent the number of one-story houses and y represent the number of two-story houses.
What are the constraints for the problem?
Mathematics
1 answer:
Annette [7]3 years ago
4 0

Answer:

5590

Step-by-step explanation:

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If I walk outside with my umbrella, then it will rain. Decide whether the given conditional statement is true or false. If it is
sergiy2304 [10]

Answer:

False

Step-by-step explanation:

Umbrellas could be used for different reasons besides rain.

4 0
3 years ago
The hypotenuse of a right triangle is 13ft long. The longer leg is 7ft longer than the shorter leg. Find the side lengths of the
DochEvi [55]

Answer:

The short side is 5ft and the longer side 12ft

Step-by-step explanation:

I knew that the hypotenuse was 13 so it couldn’t be more than 13. I was guessing at numbers until I got to 5 then added 7. After I use Pythagorean theorem to check and I got 13.

7 0
3 years ago
Read 2 more answers
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
Plzzzz help I’ve been stuck on this
alina1380 [7]

Answer:

13 and 8

Step-by-step explanation:

height is not used to find the bottom triangle

3 0
3 years ago
Find the unit vector in the direction of u = (-3,2).
DENIUS [597]
\bf \textit{unit vector for }(a,b)\implies \left( \cfrac{a}{\sqrt{a^2+b^2}}~~,~~\cfrac{b}{\sqrt{a^2+b^2}} \right)\\\\
-------------------------------\\\\
(-3,2)\qquad \stackrel{unit~vector}{\implies }\qquad \left( \cfrac{-3}{\sqrt{(-3)^2+2^2}}~~,~~\cfrac{2}{\sqrt{(-3)^2+2^2}} \right)
\\\\\\
\left( -\cfrac{3}{\sqrt{13}}~~,~~ \cfrac{2}{\sqrt{13}}\right)\\\\
-------------------------------

\bf \textit{and now let's \underline{rationalize} the denominator for each}
\\\\\\
-\cfrac{3}{\sqrt{13}}\cdot \cfrac{\sqrt{13}}{\sqrt{13}}\implies -\cfrac{3\sqrt{13}}{13} \qquad \qquad \qquad \qquad  \cfrac{2}{\sqrt{13}}\cdot \cfrac{\sqrt{13}}{\sqrt{13}}\implies \cfrac{2\sqrt{13}}{13}
\\\\\\
\textit{and written in \underline{ai+bj form}}\qquad -\cfrac{3\sqrt{13}}{13}i~~~~+~~~~\cfrac{2\sqrt{13}}{13}j
7 0
3 years ago
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