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Alexxx [7]
3 years ago
11

Please help...thanks

Mathematics
2 answers:
svetlana [45]3 years ago
6 0

Answer:

6

Step-by-step explanation:

there's 5 square + 2 halves

5+1/2+1/2=6

Andru [333]3 years ago
5 0

Answer:

6

Step-by-step explanation:

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Use slope to determine if lines PQ and RS are parallel, perpendicular, or neither P(-3,14) q(2,-1) r(4,8) s(-2,-10)
Vlad1618 [11]

Answer:

Neither

Step-by-step explanation:

The slope of PQ is -3.

The slope of RS is 3.

Parallel lines have the same slope. The slopes for perpendicular lines are the opposite of the reciprocal. PQ and RS are intersecting lines, but not parallel or perpendicular.

4 0
3 years ago
Explain why f(x) is continuous at x=3
Elina [12.6K]

Answer:

f is not  defined at x = 3 ⇒ answer (b)

Step-by-step explanation:

∵ f(x) = x² - x - 6/x² - 9 is a rational function

∴ It will be undefined at the values of x of the denominator

∵ The denominator is x² - 9

∵ x² - 9 = 0 ⇒ x² = 9 ⇒ x = ±√9

∴ x = ± 3

∴ f(x) can not be defined at x = 3

∴ The f(x) can not be continuous at x = 3

∴ The answer is (b)

6 0
3 years ago
Read 2 more answers
need helpA triangle with vertices at A(0, 0), B(0, 4), and C(6, 0) is dilated to yield a triangle with vertices at A′(0, 0), B′(
iogann1982 [59]

Answer:

2.5

Step-by-step explanation:

The dilation is 2.5. Using the point b(0,4) and B' (0,10), divide 10/4 you get 2.5.

7 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
-5/2x-19/10 <br><br> Plss Answer this in fraction form, and simplified, I can't figure it out!
Alja [10]

Answer:

= -25x-19/10

Step-by-step explanation:

8 0
3 years ago
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