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max2010maxim [7]
4 years ago
8

I need help X=? angle TGH=?

Mathematics
1 answer:
erastova [34]4 years ago
5 0

Answer:

x = 11

m<TGH = 82

Step-by-step explanation:

m<TGH + m<FGT = m<FGH

6x + 16 + 42 = 11x + 3

6x + 58 = 11x + 3

-5x = -55

x = 11

m<TGH = 6x + 16 = 6(11) + 16 = 82

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3 years ago
Find the Value of 8 1/3
Inessa [10]
8^{ \frac{1}{3} } =  \sqrt[3]{8}  = 2

8^{ \frac{2}{3} } =  \sqrt[3]{8^2} =  \sqrt[3]{64}  = 4
8 0
4 years ago
How do I solve this do we use the multiple choice
belka [17]

Answer:

  (C)  4|5 -2x| > 68

Step-by-step explanation:

You can solve each of the inequalities to see if their solutions match the given numbers, or you can start with the given numbers and see what sort of inequality you end up with.

If you plot the given "solution" on a number line, you find that the numbers -6 and 11 are the same distance from x=2.5. That distance is 8.5 units. (One way to deterimine this is to average -6 and 11, then subtract that average from 11 to find the distance.

So, we can write an inequality describing values of x that are more than 8.5 units from 2.5:

  |x -2.5| > 8.5

and it will have the solution x < -6 or x > 11.

Multiplying this by 2, it can become ...

   |2x -5| > 17

Of course, since the absolute value function doesn't care whether its argument is positive or negative, we can also write this as ...

   |5 -2x| > 17

This tells you right away which answer choice is appropriate. Further confirmation can be had by multiplying this by 4:

  4|5 -2x| > 68 . . . . . . matches selection C

5 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.

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2 years ago
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I don't know. 19? Hope this could help! ;)

4 0
3 years ago
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