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Sati [7]
2 years ago
10

Consider the function:

Mathematics
2 answers:
aivan3 [116]2 years ago
8 0

Answer:

f(x) = StartLayout enlarged left-brace 1st Row 1st column seven-halves + 2x, 2nd column x less-than-or-equal-to negative 1 2nd row 1st column negative 5 + StartFraction 3 x Over 2 EndFraction, 2nd column Negative 1 less-than x less-than 3 Third row 1st column one-fourth x, 2nd column x greater-than-or-equal-to 3 EndLayout A number line goes from negative 5 to 5.

Step-by-step explanation:

motikmotik2 years ago
4 0

Answer:

f(-3)= -5/2

f(-1) = 3/2

f(3) = 3/4

Step-by-step explanation:

edge2020

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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
Please answer!
Xelga [282]

Amswer

it is b Step-by-step explanation:

8 0
2 years ago
max lost 23 pounds while on a diet. he now weighs 184 pounds. write and solve an equation to find his initial weigh w
valina [46]

184+23=w

His initial weight is 207 pounds

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3 years ago
Change to decimals.<br> 1. 75%
ANEK [815]

Answer: 0.75

Step-by-step explanation: To write a percent as a decimal, first remember that a percent is a ratio that compares a number to 100.

So we can think of 75% as the ratio 75 to 100 or 75 ÷ 100.

Dividing by 100 moves the decimal point two places to the left so 75 ÷ 100 will move the decimal point two places to the left which would give us .75 or 0<em>.</em>75.

This means that 75% can be written as the decimal 0<em>.</em>75.

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3 years ago
What is the point (-9,-3) after rotating 180 counterclockwise
sleet_krkn [62]

Answer:

(9, 3)

Step-by-step explanation:

When a point is rotated 180 degrees, you take the opposite of the coordinates.

i.e (5, 6) ---> (-5, -6)

4 0
2 years ago
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