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andrew-mc [135]
3 years ago
14

please help asap, see attached photo. This is due in 1 hour!!! please answer both questions. Both 5 and 6

Mathematics
1 answer:
Scorpion4ik [409]3 years ago
8 0

Answer:

A and C

Step-by-step explanation:

5) logx+log(x+9)=2

log(x(x+9))=2, log(x^2+9x)=2

6) log(x^2+9x)=2

(x^2+9x)=36 or 6^2

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Simplify completely quantity 12 x plus 36 over quantity x squared minus 4 x minus 21 and find the restrictions on the variable..
Softa [21]
To simplify the expression stated, it is better to write it in the numerical form. The statement is:

<span>quantity 12 x plus 36 over quantity x squared minus 4 x minus 21

The numerical form is:

12x+36 / (x</span>²-4x-21)

Simplifying,
12x+36 / (x<span>²-4x-21)
</span>12 (x+3) / (x+3)(x-7)

We can cancel the x+3 term, leaving only:
12/(x-<span>7)</span>
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3 years ago
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Which function has a vertex on the y-axis? f(x) = (x – 2)2 f(x) = x(x + 2) f(x) = (x – 2)(x + 2) f(x) = (x + 1)(x – 2)
strojnjashka [21]

we know that

If the vertex is on the y-axis, then the x-coordinate of the vertex is equal to zero

we are going to verify the vertex of each one of the functions to determine the solution

Remember that

The equation in vertex form of a vertical parabola is equal to

y=a(x-h)^{2} +k

where

(h,k) is the vertex

if a>0 -------> the parabola open upward (vertex is a minimum)

if a -------> the parabola open downward (vertex is a maximun)

<u>case A)</u> f(x)=(x-2)^{2}

This is a vertical parabola open upward

the vertex is the point (2,0)

therefore

The function f(x)=(x-2)^{2}  does not have a vertex on the y-axis

<u>case B)</u> f(x)=x(x+2)

f(x)=x(x+2)=x^{2}+2x

convert to vertex form

Complete the square. Remember to balance the equation by adding the same constants to each side.

f(x)+1=x^{2}+2x+1

Rewrite as perfect squares

f(x)+1=(x+1)^{2}

f(x)=(x+1)^{2}-1

the vertex is the point (-1,-1)

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The function f(x)=x(x+2) does not have a vertex on the y-axis

<u>case C)</u> f(x)=(x-2)(x+2)

f(x)=(x-2)(x+2)=x^{2}-2^{2}

f(x)=x^{2}-4

the vertex is the point (0,-4)

The x-coordinate of the vertex is equal to zero

therefore

The function f(x)=(x-2)(x+2) has a vertex on the y-axis

<u>case D)</u> f(x)=(x+1)(x-2)

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

convert to vertex form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

f(x)+2= x^{2} -x

Complete the square. Remember to balance the equation by adding the same constants to each side.

f(x)+2+0.25= x^{2} -x+0.25

f(x)+2.25= x^{2} -x+0.25

Rewrite as perfect squares

f(x)+2.25= (x-0.50)^{2}

f(x)=(x-0.50)^{2}-2.25

the vertex is the point (0.5,-2.25)

therefore

The function f(x)=(x+1)(x-2) does not have a vertex on the y-axis

<u>the answer is</u>

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

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Graph 3x+2y=4.
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In two or more complete sentences, define the key features of the graph below and explain how to write its equation using functi
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The graph has a vertex at (3, -2). It extends upward from there linearly at a slope of -1 to the left and 1 to the right. It is the graph of an absolute value function. If we assume it keeps extending upwards the domain is all real numbers. (which is what i would assume even though there's no arrows it doesn't have decipherable endpoints). The range is y ≥ -2 with y -intercept (0,1), and x-intercepts: (5,0) & (1,0).

To write the equation for this function, I would acknowledge that it is the translation of the graph of the standard absolute value function: f(x) = |x| ; right 3 and down 2. Which would be to subtract 3 from x and subtract 2 from the end.
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Neko [114]

Answer:

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