All categories

3 years ago

Plzzzzz helpmeeeee plzzzzzz

Mathematics

1 answer:

Nutka1998 [239]3 years ago

5 0

C is your answer in figures. one million and one hundred and ten thousand dollars

You might be interested in

It took the Phillips family 6 hours

Lerok [7]

Answer:

Percentage increase.

Step-by-step explanation:

It took the Phillips family 6 hours to drive to their vacation destination. 2 hours and 42 minutes extra because of traffic.

8 hours 42 minutes

6 hours 0 minutes

2 hours and 42 minutes

Percentage increase.

8 0

2 years ago

Problem in the attachment

TEA [102]

You can solve this by multiplying each numerator by the answer choices and trying to divide them. I can't get the fraction maker to work at the moment, so I'll write the calculations on a separate attachment.

As you can see, the only choice that works to eliminate both fractions is the fourth choice, 50x.

8 0

3 years ago

What is the coordinate of the image of point (5,8) after it is translation 7 units to the left units up

Whitepunk [10]

(-1,x) you didnt say how many units up

6 0

3 years ago

Suppose the bank installs nine alarms. What is the expected number of alarms that will detect a burglar?

Harrizon [31]

The answer is 6 because

3 0

3 years ago

19) Given that f(x)x² - 8x+ 15x² - 25find the horizontal and vertical asymptotes using the limits of the function.A) No Vertical

Tems11 [23]

EXPLANATION

Since we have the function:

$f(x)=\frac{x^2-8x+15}{x^2}$

Vertical asymptotes:

$For\:rational\:functions,\:the\:vertical\:asymptotes\:are\:the\:undefined\:points,\:also\:known\:as\:the\:zeros\:of\:the\:denominator,\:of\:the\:simplified\:function.$

Taking the denominator and comparing to zero:

$x+5=0$

The following points are undefined:

$x=-5$

Therefore, the vertical asymptote is at x=-5

Horizontal asymptotes:

$\mathrm{If\:denominator's\:degree\:>\:numerator's\:degree,\:the\:horizontal\:asymptote\:is\:the\:x-axis:}\:y=0.$ $If\:numerator's\:degree\:=\:1\:+\:denominator's\:degree,\:the\:asymptote\:is\:a\:slant\:asymptote\:of\:the\:form:\:y=mx+b.$ $If\:the\:degrees\:are\:equal,\:the\:asymptote\:is:\:y=\frac{numerator's\:leading\:coefficient}{denominator's\:leading\:coefficient}$ $\mathrm{If\:numerator's\:degree\:>\:1\:+\:denominator's\:degree,\:there\:is\:no\:horizontal\:asymptote.}$ $\mathrm{The\:degree\:of\:the\:numerator}=1.\:\mathrm{The\:degree\:of\:the\:denominator}=1$ $\mathrm{The\:degrees\:are\:equal,\:the\:asymptote\:is:}\:y=\frac{\mathrm{numerator's\:leading\:coefficient}}{\mathrm{denominator's\:leading\:coefficient}}$ $\mathrm{Numerator's\:leading\:coefficient}=1,\:\mathrm{Denominator's\:leading\:coefficient}=1$ $y=\frac{1}{1}$ $\mathrm{The\:horizontal\:asymptote\:is:}$ $y=1$

In conclusion:

$\mathrm{Vertical}\text{ asymptotes}:\:x=-5,\:\mathrm{Horizontal}\text{ asymptotes}:\:y=1$

4 0

1 year ago