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VladimirAG [237]

3 years ago

7

Amy started with $125 in her savings account. She deposited $25, withdrew $15, and then deposited $30. How much money is in Amy’

s savings account now?

2 answers:

oksian1 [2.3K]3 years ago

5 0

Answer:

she now has $165 in her savings account

Leya [2.2K]3 years ago

5 0

Answer:

$165

Step-by-step explanation:

125+25=150

150-15=135

135+30=165

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Vesnalui [34]

Answer:

7/0

Step-by-step explanation:

This is because if a number is divided by 0 then there is no answer or it is undefined

Think of it like this,

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2 years ago

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The graph shows g(x), which is a translation of f(x) = x². Write the function rule for g(x).

mixas84 [53]

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Step-by-step explanation:

7 0

2 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