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liberstina [14]
2 years ago
12

Determine the x-intercept(s) of the rational function: f(x) = (x ^ 2 - 16)/(x ^ 2 - 2x - 3)

Mathematics
1 answer:
elena55 [62]2 years ago
3 0

Answer:

(-4,0) and (4,0)

Step-by-step explanation:

The x-intercepts occur for values of x where f(x) = 0.

f(x) = 0 when the nominator (x^{2}  - 16) = 0.

x^{2} -16 is a difference of squares,
so you can factor it as (x + 4)(x – 4)

When x = –4 or x = 4, the nominator is 0, and f(x) = 0

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SSSSS [86.1K]

Answer:

Angles 6, 4, and 2 are congruent to angle 8.

Step-by-step explanation:

By the Opposite Angles Theorem, angle 6 is congruent to angle 8.

By the Corresponding Angles Theorem, angle 4 is congruent to angle 8.

By the Corresponding Angles Theorem, angle 2 is congruent to angle 6, and angle 6 is congruent to angle 8, so angle 2 is congruent to angle 8 by the Transitive Property.

7 0
2 years ago
What is 19 and 1/3 multiplied by 5 and 4/5
Nadusha1986 [10]

Answer:

112 2/15

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Determine if the lines r1(t) = <3, 0, 2> + t <1, 2, −2> and r2(s) = <0, 1, −1> + s <4, 1, 1> intersect,
Temka [501]

Answer:

Yes lines are intersecting, point of intersection is <4,2,0>.

Step-by-step explanation:

Given parametric equations of line are:

r_{1}(t) =  + t  \\ r_{1}(t) = \\ r_{1}(t) = ---(1)

r_{2}(s) =  + s \\ r_{2}(s) = \\ r_{2}(s) = ---(2)\\

If lines are intersecting then parametric coordinates of (1) are equal to (2)

3+t=4s---(A)\\2t=1+s---(B)\\2-2t=s-1---(C)\\

Considering A and B to find values of t and s

From A

t=4s-3---(D)

Putting in (B)

2(4s-3)=1+s

8s-6=1+s

7s=7

s=1

Then

t=4-3

t=1

If lines are intersecting then these values of s and t must satisfy (C)

2-2(1)=1-1

0=0

This shows lines are intersecting.

At this value of t, (1) becomes

r_{1}(1) = \\=

Putting s=1 in (2)

r_{2}(1)=4, 1+1,-1+1>\\=

Point of intersection is <4,2,0>.

4 0
3 years ago
Please help me i need it right now please
12345 [234]

Given:

The rate of interest on three accounts are 7%, 8%, 9%.

She has twice as much money invested at 8% as she does in 7%.

She has three times as much at 9% as she has at 7%.

Total interest for the year is $150.

To find:

Amount invested on each rate.

Solution:

Let x be the amount invested at 7%. Then,

The amount invested at 8% = 2x

The amount invested at 9% = 3x

Total interest for the year is $150.

x\times \dfrac{7}{100}+2x\times \dfrac{8}{100}+3x\times \dfrac{9}{100}=150

Multiply both sides by 100.

7x+16x+27x=15000

50x=15000

Divide both sides by 50.

x=\dfrac{15000}{50}

x=300

The amount invested at 7% is 300.

The amount invested at 8% is

2(300)=600

The amount invested at 9% is

3(300)=900

Thus, the stockbroker invested $300 at 7%, $600 at 8%, and $900 at 9%.

4 0
3 years ago
I need help with proportions and this question, if you can help thank you.
DanielleElmas [232]
Ok give me a second.
3 0
3 years ago
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