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klasskru [66]
3 years ago
8

A wall in marcus room is 8 2/5 feet high as 16 2/3 feet long. If he paints 1/2 the wall blue how many square feet will be blue

Mathematics
1 answer:
svlad2 [7]3 years ago
4 0
The answer is 70. Hope this helps!
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Helpp with this math.
Annette [7]

Answer:

it's c lol

Step-by-step explanation:

5 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
8x + y 17<br> 4x + 3y - 31
Vladimir [108]

1) 8x+y\cdot \:17:\quad 8x+17y

2) 4x+3y-31:\quad 4x+3y-31

7 0
2 years ago
Read 2 more answers
12x^2+20=50 what is x
LenaWriter [7]

Answer: x=1.5811389999999999 , or 1.59

Step-by-step explanation:

First, Subtract 20 from both sides.

      12x2+20−20=50−20

      12x2=30

 Then, Divide both sides by 12:

12x^2/12 = 30/12

⇒ you get : x^2 = 5/2

After, Take square root.

x= ± √ 5/2

Finally you answer is going to be =1.5811389999999999 OR, 1.59

* Hopefully this helps:) Mark me the brainliest:)

<em>∞ 234483279c20∞</em>

7 0
3 years ago
Units of Capacity
marysya [2.9K]

Answer:

            \large\boxed{\large\boxed{0.2deciliter}}

Explanation:

Build your conversion factors using the conversion table:

  • 1 cup = 0.237 liters\implies 1=0.237liters/1cup

From the prefixes used in the metric system, you know that 1 deciliter = 0.100 liters. Then:

  • 1 deciliter = 0.100 liters\implies 1=1deciliter/0.100liters

Arrange the expression to simplify the units and convert from cups to deciliters:

  • 9cups\times 0.237liters/cup\times 1deciliter/0.100liters=0.2133deciliter

Round to the nearest tenth: 0.2 deciliter.

4 0
3 years ago
Read 2 more answers
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