All categories

2 years ago

The temperature was -7 degrees in the morning. By noon, it was 12 degrees outside. What was the change in temperature?

Mathematics

1 answer:

11111nata11111 [884]2 years ago

6 0

Answer:

19 degrees

Step-by-step explanation:

12-(-7)= 19

(12+7=19)

You might be interested in

35POINTS PLZ HELP FAST

meriva

Answer:

Step-by-step explanation:

16.76FT^3

3 0

2 years ago

Find the first five terms of each geometric sequence described A1=1 R=4

xenn [34]

$\bf n^{th}\textit{ term of a geometric sequence}
\\\\\\
a_n=a_1\cdot r^{n-1}\qquad 
\begin{cases}
r=\textit{common ratio}\\
a_1=\textit{first term}\\
-------\\
a_1=1\\
r=4
\end{cases}\\\\
-----------------------------\\\\
$

$\bf \begin{array}{ccllll}
term&value\\
\textendash\textendash\textendash\textendash\textendash\textendash&\textendash\textendash\textendash\textendash\textendash\textendash\\
1&a_1=1\\
2&a_2=a_1\cdot 4^{2-1}\\
3&a_3=a_1\cdot 4^{3-1}\\
4&a_4=a_1\cdot 4^{4-1}\\
5&a_5=a_1\cdot 4^{5-1}\\
\end{array}$

3 0

3 years ago

Solve for d. 2d+3=7d+18

algol [13]

Answer: d = -3

Step-by-step explanation:

2d+3=7d+18

-5d=15

d=-3

5 0

2 years ago

Read 2 more answers

Identify the initial value and rate of change for the graph shown. (4 points)

spin [16.1K]

Answer:

Step-by-step explanation:

8 0

3 years ago

*Asymptotes* g(x) =2x+1/x-3 Give the domain and x and y intercepts

Nataly [62]

Answer: Assuming the function is $g(x)=\frac{2x+1}{x-3}$ :

The x-intercept is $(\frac{-1}{2},0)$ .

The y-intercept is $(0,\frac{-1}{3})$ .

The horizontal asymptote is $y=2$ .

The vertical asymptote is $x=3$ .

Step-by-step explanation:

I'm going to assume the function is: $g(x)=\frac{2x+1}{x-3}$ and not $g(x)=2x+\frac{1}{x}-3$ .

So we are looking at $g(x)=\frac{2x+1}{x-3}$ .

The x-intercept is when y is 0 (when g(x) is 0).

Replace g(x) with 0.

$0=\frac{2x+1}{x-3}$

A fraction is only 0 when it's numerator is 0. You are really just solving:

$0=2x+1$

Subtract 1 on both sides:

$-1=2x$

Divide both sides by 2:

$\frac{-1}{2}=x$

The x-intercept is $(\frac{-1}{2},0)$ .

The y-intercept is when x is 0.

Replace x with 0.

$g(0)=\frac{2(0)+1}{0-3}$

$y=\frac{2(0)+1}{0-3}$

$y=\frac{0+1}{-3}$

$y=\frac{1}{-3}$

$y=-\frac{1}{3}$ .

The y-intercept is $(0,\frac{-1}{3})$ .

The vertical asymptote is when the denominator is 0 without making the top 0 also.

So the deliminator is 0 when x-3=0.

Solve x-3=0.

Add 3 on both sides:

x=3

Plugging 3 into the top gives 2(3)+1=6+1=7.

So we have a vertical asymptote at x=3.

Now let's look at the horizontal asymptote.

I could tell you if the degrees match that the horizontal asymptote is just the leading coefficient of the top over the leading coefficient of the bottom which means are horizontal asymptote is $y=\frac{2}{1}$ . After simplifying you could just say the horizontal asymptote is $y=2$ .

Or!

I could do some division to make it more clear. The way I'm going to do this certain division is rewriting the top in terms of (x-3).

$y=\frac{2x+1}{x-3}=\frac{2(x-3)+7}{x-3}=\frac{2(x-3)}{x-3}+\frac{7}{x-3}$

$y=2+\frac{7}{x-3}$

So you can think it like this what value will y never be here.

7/(x-3) will never be 0 because 7 will never be 0.

So y will never be 2+0=2.

The horizontal asymptote is y=2.

(Disclaimer: There are some functions that will cross over their horizontal asymptote early on.)

6 0

3 years ago