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tatuchka [14]
3 years ago
11

Y=x+6x+y=8now graph 3 points on the line x+y=8​

Mathematics
1 answer:
Brums [2.3K]3 years ago
8 0

Answer:

y=2

Step-by-step explanation:

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G(x) = 3x + 1; Find g(-8)
gregori [183]

Answer:

g(-8) = -23

Step-by-step explanation:

g(x) = 3x + 1    <em>Plug in g(-8)</em>

g(-8) = 3(-8) + 1  <em>Multiply 3 by -8</em>

g(-8) = -24 + 1   <em>Add 1 to -24</em>

g(-8) = -23

5 0
3 years ago
Read 2 more answers
An equation of an ellipse is given. x2 16 + y2 25 = 1 (a) Find the vertices, foci, and eccentricity of the ellipse. vertex (x, y
eimsori [14]

Answer:

(a)

  • The vertices are at (0,-5) and (0,5).
  • The coordinates of the foci are (0,-3) and (0,3).
  • Eccentricity=3/5

(b)Length of the major axis=10

Step-by-step explanation:

When the major axis of an ellipse is parallel to the y-axis.The standard form of the equation of an ellipse is given as:

\dfrac{x^2}{b^2}+\dfrac{y^2}{a^2}=1

Given the equation:

\dfrac{x^2}{16}+\dfrac{y^2}{25}=1

(I)The coordinates of the vertices are (0, \pm a)

a^2=25\\a^2=5^2\\a=5

Therefore, the vertices are at (0,-5) and (0,5).

(II)The coordinates of the foci are (0, \pm c)$ where c^2=a^2-b^2

c^2=a^2-b^2\\c^2=25-16\\c^2=9\\c=3

The coordinates of the foci are (0,-3) and (0,3).

(III)Eccentricity

This is the ratio of the distance c between the center of the ellipse and each focus to the length of the semi major axis.

Simply put, Eccentricity =c/a

Eccentricity=3/5

(b)Length of the major axis

The length of the major axis=2a

=2(5)=10.

4 0
3 years ago
The graph below shows the function f(x)=x-3/x2-2x-3. Which statement is true?
BARSIC [14]
1. Domain.

We have x^2-2x-3 in the denominator, so:

x^2-2x-3\neq0\\\\(x^2-2x+1)-4\neq0\\\\(x-1)^2-4\neq0\\\\(x-1)^2-2^2\neq0\qquad\qquad[\text{use }a^2-b^2=(a-b)(a+b)]\\\\(x-1-2)(x-1+2)\neq0\\\\&#10;(x-3)(x+1)\neq0\\\\\boxed{x\neq3\qquad\wedge\qquad x\neq-1}

So there is a hole or an asymptote at x = 3 and x = -1 and we know, that answer B) is wrong.

2. Asymptotes:

f(x)=\dfrac{x-3}{x^2-2x-3}=\dfrac{x-3}{(x-3)(x+1)}=\boxed{\dfrac{1}{x+1}}

We have only one asymptote at x = -1 (and hole at x = 3), thus the correct answer is A)
6 0
3 years ago
Read 2 more answers
Solve for x. Round to the nearest tenth.
Shkiper50 [21]

Answer:

24.8

Step-by-step explanation:

5 0
2 years ago
Find f(-2) for the function:<br><br> f(x)= 3x^2-1, x&lt;1<br> x+2, x&gt; 1
azamat
F(x) = 3x² - 1
f(-2) = 3(-2)² - 1
f(-2) = 3(4) - 1
f(-2) = 12 - 1
f(-2) = 11

f(x) = x < 1
f(-2) = -2 < 1

f(x) = x + 2
f(-2) = -2 + 2
f(-2) = 0

f(x) = x > 1
f(-2) = -2 > 1
f(-2) = -2 < 1
8 0
3 years ago
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