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boyakko [2]
3 years ago
5

Which expression is equivalent to 4.4.4.4.4.4.4.4?

Mathematics
2 answers:
seraphim [82]3 years ago
8 0

Answer:

your answer is D

Step-by-step explanation:

hope this helps

AlexFokin [52]3 years ago
5 0
The last answer is the correct answer
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Need help with these problems and I have to show my work
Rasek [7]

Answer:

1. 20 min and 15 sec

2. 4 min and 30 sec

3. y=80x-20

4. the second one

Step-by-step explanation:

1.  in 2 min you have 25 g

in 15 sec you loose 2.5 g

2. in 4 min you have 5 g

in 30 sec you loose 5 g

3. y=45x-10

   y=80x-20

4. it takes 2 min and 15 sec to empty the first one and 3 min to empty the second one.

8 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
Sombody please help me out.​
Nata [24]

Answer:

B. (x, y) --> (x, y + 5)

Step-by-step explanation:

Y is up and down

X is left and right

So assuming that Y is north she needs to go 5 units up Y

(x, y + 5) = (x, 5y)

4 0
3 years ago
Which is the graph of the linear inequality y &lt; 3x + 1?<br> 3<br> 1<br> 3<br> 4.<br> 5
Arlecino [84]

Answer:

Image below

Step-by-step explanation:

<em>Hey there!</em>

<em />

Well lets graph the given inequality,

y < 3x + 1

Look at the image below.

So the right graph should look like the image below.

<em>Hope this helps :)</em>

6 0
3 years ago
I need help answering questions like this can someone help
Leviafan [203]

Answer:

300 in.³

Step-by-step explanation:

We find the volume of 1 prism and multiply it by 2.

V = BH

where V = volume of prism, B = area of the base of the prism, H = height of the prism

V = ½bh × H

V = ½(5 in.)(6 in.) × 10 in.

V = 150 in.³

That is the volume of one triangular prism. Now we multiply it by 2.

volume = 2 × 150 in.³ = 300 in.³

4 0
3 years ago
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