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3 years ago

ANSWER QUICK PLZ

Mathematics

2 answers:

alexandr402 [8]3 years ago

6 0

Answer:

at least one x-intercept

at least one y-intercept

a vertical asymptote

the domain

Step-by-step explanation:

g(x)=(-4x*2+36)/(x-3)

g(x) = -4(x² - 9)/(x - 3)

g(x) = -4(x - 3)(x + 3)/(x - 3)

g(x) = -4(x + 3)

With a hole at x = 3

Bad White [126]3 years ago

3 0

Answer:

see below

Step-by-step explanation:

at least one x-intercept - where it crosses the x axis- They both have at least one x intercept

at least one y-intercept - where it crosses the y axis- they both have at least one

an oblique asymptote or slant asymptote- when the degree of the polynomial in the numerator is higher than the denominator - this is true

a vertical asymptote - they both have a vertical asymptote

the domain - is all real values except x = -3 in one case and x=3 in the other

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(05.03 LC)

Paladinen [302]

$f(x)=3x-23,\ g(x)=-4.5x+7\\\\f(x)=g(x)\iff3x-23=-4.5x+7\qquad\text{add 23 to both sides}\\\\3x=-4.5x+30\qquad\text{add 4.5x to both sides}\\\\7.5x=30\qquad\text{divide both sides by 7.5}\\\\\boxed{x=4}$

6 0

3 years ago

Read 2 more answers

Although cities encourage carpooling to reduce traffic congestion, most vehicles carry only one person. For example, 64% of vehi

ira [324]

Answer:

a) 0.7291 is the probability that more than half out of 10 vehicles carry just 1 person.

b) 0.996 is the probability that more than half of the vehicles carry just one person.

Step-by-step explanation:

We are given the following information:

A) Binomial distribution

We treat vehicle on road with one passenger as a success.

P(success) = 64% = 0.64

Then the number of vehicles follows a binomial distribution, where

$P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}$

where n is the total number of observations, x is the number of success, p is the probability of success.

Now, we are given n = 10

We have to evaluate:

$P(x \geq 6) = P(x =6) +...+ P(x = 10) \\= \binom{10}{6}(0.64)^6(1-0.64)^4 +...+ \binom{10}{10}(0.64)^{10}(1-0.79)^0\\=0.7291$

0.7291 is the probability that more than half out of 10 vehicles carry just 1 person.

B) By normal approximation

Sample size, n = 92

p = 0.64

$\mu = np = 92(0.64) = 58.88$

$\sigma = \sqrt{np(1-p)} = \sqrt{92(0.64)(1-0.64)} = 4.60$

We have to evaluate the probability that more than 47 cars carry just one person.

$P(x \geq 47)$

After continuity correction, we will evaluate

$P( x \geq 46.5) = P( z > \displaystyle\frac{46.5 - 58.88}{4.60}) = P(z > -2.6913)$

$= 1 - P(z \leq -2.6913)$

Calculation the value from standard normal z table, we have,

$P(x > 46.5) = 1 - 0.004 = 0.996 = 99.6\%$

0.996 is the probability that more than half out of 92 vehicles carry just one person.

5 0

3 years ago

PlEASE HELP URGENT!!!!!!!!!

ohaa [14]

The answer is -3, i had a test with the same question

8 0

2 years ago

Read 2 more answers

Find the distance between each pair points (-5,3) , (-4,-5)?

tester [92]

The distance formula is d=√(x2-x1)² + (y2-y1)²

So... d =8.06226

Hope this helps

7 0

3 years ago

What values complete each statement?

aliya0001 [1]

For this case we have:

By properties of the radicals $\sqrt {a} = a ^{\frac {1} {2}}$

So:

$(\sqrt {7}) ^ 2 = (7 ^ {\frac {1} {2}}) ^ 2$ .

Now, for power properties we have:

$(b ^ {\frac {c} {d}}) ^ e = b ^ {\frac {c * e} {d}}$

Thus, $(7 ^ {\frac {1} {2}}) ^ 2 = 7 ^ {\frac {2} {2}} = 7$

So:

$7 ^ {\frac {1} {2}} = \sqrt {7}$ in its radical form

Answer:

$(\sqrt {7}) ^ 2 = (7 ^ {\frac {1} {2}}) ^ 2= 7 ^ {\frac {2} {2}} = 7$ in its simplest form.

$7 ^ {\frac {1} {2}} = \sqrt {7}$ in its radical form

4 0

3 years ago