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antiseptic1488 [7]
3 years ago
10

Write an equation of the line with a slope of 2/3 and y-intercept of -8.

Mathematics
1 answer:
Zolol [24]3 years ago
4 0

Answer:

y = mx + b

y = 2/3x + -8

M = 2/3

y -intercept b is -8

Step-by-step explanation:

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Evaluate 8P4 how can I evaluate this problem
Leno4ka [110]
NPr = n! / (n-r)!

8P4 = 8! / (8-4)!

= 8! / 4!

= 1,680
5 0
3 years ago
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1) Choose all of the common denominators of 2/3 and 7/9.
pishuonlain [190]

Common denominators of 9 and 3 are 9. 27

Common denominators of 9 and 2 are 18 36

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
Can you please help me thanks
mash [69]

Answer:

x = 9.6

Step-by-step explanation:

Before we can figure out what x is, we need to figure out what the unlabeled side is. To figure that out, multiply the hypotenuse (12) by the sine of the angle labeled (34)

12 * sin(34)

<em>sin(34) equals 0.559192903470747; I'll round it to 0.6 for convenience.</em>

12 * 0.6

<em>Now simply multiply 12 by 0.6 to get 7.2.</em>

The unlabeled side is approx. 7.2 units long.

Now we know what the unlabeled side is. Now, to find x, find the square root of 12 squared minus 7.2 squared.

x = √12² - 7.2²

<em>12 squared is 144; 7.2 squared is 51.84.</em>

x = √144 - 51.84

<em>Subtract 51.84 from 144 to get 92.16.</em>

x = √92.16

<em>The square root of 92.16 is 9.6 (on the spot!).</em>

x equals 9.6.

6 0
3 years ago
What is the volume, in cubic cm, of a rectangular prism with a height of 19cm, a
3241004551 [841]

Answer: 2432

19 x 16 x 8 = 2432

6 0
3 years ago
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