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VikaD [51]
3 years ago
8

Which of the following are possible values for x?

Mathematics
1 answer:
Natali5045456 [20]3 years ago
4 0

Answer:

The answer is only c) x=5

Step-by-step explanation:

|x| means that this is an absolute value. If the number is a negative, removing the value bars will make it a positive again. Positives stay positives. So the only answer greater than value of 2 is 5.

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Help me please, please help
Sergio [31]

Answer:

Heyo. The answer is 60.48

Step-by-step explanation:

This is pretty easy, all you have to do is multiply the length, height, and width. Remember this. LxHxW. Thats is the formula. You do 4.2 cm, multiplied by 3.2 cm, then multiply that by 4.5 cm. Hope this helps! :)

8 0
3 years ago
SOMEOne PLS EXPLAIN VERTICAL ASYMPTOTES TO ME IM DYING- i need to know how u get the asymptote(explanation) and answer T^TTTTTTT
Kitty [74]
A vertical asymptote is what you get when you try to divide by 0. To find where you get these, you need to look at the denominator and what values of x will make the denominator equal to 0.

In your denominator, you have (x+7)(x-5)(x-3).
What values of x makes (x+7)(x-5)(x-3)=0?
If x = -7, if x = 5, or if x = 3, then that entire expression will equal zero. (Same idea as when you solve equations by factoring.

Now the only place this can get trickier is if one of those factors — one of (x+7), (x-5), or (x-3) — also appears in the numerator. If that happens, then it’s more involved whether you have an asymptote or not. But that doesn’t happen in this example.

So the short version: Asymptotes happen when you try to divide by zero. Dividing by zero is not a good thing. So you just ask yourself, “What will make the denominator 0?”
4 0
2 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
C. 24 less then double x<br> .
beks73 [17]

Answer:

For this type of question you might want to give a domain in a set statement

ie.24 is less than 2(13,14,15,...)

<em><u>Or</u></em>

<em>If</em><em> </em><em>my</em><em> </em><em>fi</em><em>rst</em><em> </em><em>answer</em><em> </em><em>was</em><em> </em><em>a</em><em> </em><em>deviation</em><em> </em><em>then</em><em> </em><em> you</em><em> </em><em>were</em><em> </em><em>proba</em><em>bly</em><em> </em><em>being</em><em> </em><em>ask</em><em>ed</em><em> to</em><em> </em><em>subt</em><em>ract</em><em> </em><em>2</em><em>4</em><em> </em><em>fro</em><em>m</em><em> </em><em>2</em><em>t</em><em>i</em><em>m</em><em>e</em><em>s</em><em> </em><em>a</em><em> </em><em>giv</em><em>en</em><em> </em><em>num</em><em>ber</em><em> </em><em>being</em><em> </em><em>repres</em><em>e</em><em>nted</em><em> </em><em> </em><em>by</em><em> </em><em>the</em><em> </em><em>varia</em><em>ble</em><em> </em><em>x</em>

6 0
3 years ago
if x is the first, or smallest, of three consecutive integers, express the sum of the integer and the third integer as an algebr
Scrat [10]

Consecutive numbers differ by one. So, if x is the first, the three numbers are

x,\ x+1,\ x+2

The sum of the first and third integer is

x+(x+2)=2x+2

5 0
3 years ago
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