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2 years ago

This is due today !! help pls

Mathematics

2 answers:

jeka57 [31]2 years ago

8 0

Answer:

you are going to do angles sum on 180 so what you are to do is 5+12+B+c=180

Anarel [89]2 years ago

5 0

Answer:

Step-by-step explanation:

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Select the correct answer.

Ivan

Diagram needed please!!

7 0

3 years ago

Read 2 more answers

Let f(x)=(18)3^x. find the equation of the line between the points (-2,f(-2)) and (1,f(1))

horsena [70]

$\text{Given that,} \\\\f(x) = 18 \cdot 3^x \\\\f(-2) = 18 \cdot 3^{-2} = 2\\\\f(1) = 18 \cdot 3^1 = 54\\ \\\text{So,}~ (x_1,y_1) = (x_1, f(-2)) = (-2,2) ~\text{and}~ (x_2,y_2) = (x_2, f(1)) = (1,54)\\ \\\text{Slope,}~ m = \dfrac{y_2 -y_1}{x_2 -x_1} = \dfrac{54-2}{1+2} = \dfrac{52}3 \\\\$

$\text{Equation of line,}\\\\~~~~~~~y - y_1 = m (x-x_1)\\\\\\\implies y-2 = \dfrac{52}3(x+2)~~~~~~~~~~~;[\text{Point-Slope form.]} \\\\\\\implies y-2 = \dfrac{52}3x + \dfrac{104}3\\\\\\\implies y = \dfrac{52}{3}x + \dfrac{104} 3 +2\\\\\\\implies y = \dfrac{52}3 x +\dfrac{110}3 ~~~~~~~~~~~~~~~;[\text{Slope -Intercept form}]$

6 0

2 years ago

If y=3x+ 2 were changed to y=3x+8 how would the graph of the new function compare with the first one?

Alex787 [66]

D)it would be shifted up

3 0

3 years ago

If 17 is 32 than what does 17+17+17+17+17=?

IceJOKER [234]

Answer:

The answer is 160

Step-by-step explanation:

Comment

There really should be no answer to this. It sort of destroys the meaning of equality. However, you might be working in different bases. I'll assume that, but the base will be a fraction, which brings up other problems.

Solution

You have 5 17s there. So you will need five 32s. 5 * 32 = 160

The answer is 160

5 0

1 year 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