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Anit [1.1K]
4 years ago
9

How do I graph quadratic equations: y=(x+2)^2-1

Mathematics
1 answer:
Fudgin [204]4 years ago
6 0
Easiest way is if you have a graphing calculator, but if not. Simply plug in values for x, starting at -2 for this equation, and then find the corresponding y values. Then plot.
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Use the Distributive Property to<br> REWRITE -2(4 - 7x)
Yuri [45]

Answer:

-8+14x

Step-by-step explanation:

By distribution, -2(4-7x) = -8+14x

5 0
3 years ago
<img src="https://tex.z-dn.net/?f=8%20%5E%28x%29%2B32%20x%5E%7B2%7D%2B4" id="TexFormula1" title="8 ^(x)+32 x^{2}+4" alt="8 ^(x)+
Grace [21]

Answer:

Step-by-step explanation:

Rearrange the terms and take out a common factor of 4

4(8x^2 + 2x + 1)

I think this is as far down as you want to take it. It gives an imaginary root which you may not be familiar with.

4 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
2 years ago
If T is the midpoint of CE then CT=TE
mr_godi [17]
Answer: true statement
6 0
4 years ago
−7 + 3(8n − 8) = 8(7 + 3n) please show work
stepladder [879]

Answer:

Step 1: Simplify both sides of the equation.

−7+3(8n−8)=8(7+3n)

−7+(3)(8n)+(3)(−8)=(8)(7)+(8)(3n)(Distribute)

−7+24n+−24=56+24n

(24n)+(−7+−24)=24n+56(Combine Like Terms)

24n+−31=24n+56

24n−31=24n+56

Step 2: Subtract 24n from both sides.

24n−31−24n=24n+56−24n

−31=56

Step 3: Add 31 to both sides.

−31+31=56+31

0=87

Answer:

There are no solutions.

Step-by-step explanation:

3 0
3 years ago
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