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4 years ago

How do you write the slope and the y-intercept of the line

Mathematics

1 answer:

Readme [11.4K]4 years ago

4 0

Answer:

The equation is y=mx+b, where m= slope and b= y-intercept

If your slope was 2 and your intercept was 3, it would look like:

y= 2x+3

Step-by-step explanation:

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Use the given graph. Determine the period of the function.

Ivahew [28]

Answer:

Step-by-step explanation:

Period of a function is the period after which the function attains the same value

in the graph attached with this problem we can see that

f(0)=1

the value of x for which function f(x) attains the value 1 again is at

x=3

f(3)=1

similarly , we see

f(6)=1 , f(9)=1

Hence we see that after every increased value of x by 3 units , we attain the same value of function . hence the period of the function is 3

6 0

3 years ago

Matt wants to play in the sand pit, and then try the swings, the slide, and the seesaw, in that order. If the units on the coord

Nataly [62]

Answer:

it should be 12

i think :)

i hope im correct

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Make n the subject of the formula

kogti [31]

Answer:

$\mathsf{ \frac{m + 21}{5} = n }$

Step-by-step explanation:

To make "n" the subject of the formula, rearrange the formula so it begins with " n = "

To isolate the variable "n", you need to inverse the other terms on that side of the equation where "n" really is.

=> m = 5n - 21

there's a "-21" next to "5n", inverse of subtraction is addition, so add 21 on both sides of the equation.

=> m + 21 = 5n - 21 + 21

=> m+ 21 = 5n

There's also a 5 attached to n, that means the multiplication of n with 5, the inverse of multiplication is division, so dividing both sides of the equation by 5

$\mathsf{ \frac{m + 21}{5} = n }$

3 0

3 years ago

Suppose you invest $400 at an annual interest rate 7.6% compounded continuously. How much will you have in the account after 1.5

goldfiish [28.3K]

$\bf ~~~~~~ \textit{Continuously Compounding Interest Earned Amount}\\\\
A=Pe^{rt}\qquad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to& \$400\\
r=rate\to 7.6\%\to \frac{7.6}{100}\to &0.076\\
t=years\to &1.5
\end{cases}
\\\\\\
A=400e^{0.076\cdot 1.5}\implies A=400e^{0.114}$