Someone PLease please help me

Write the point-slope form of the equation of the line passing through the points (-5, 6) and (0, 1).

andrey2020 [161]

Formula of the slope of a linear function:

m = (y₂ - y₁)/(x₂ - x₁)

m= (-1 - 6)/[0- (-5)]

m= -7/5 This is the slope requested

7 0

3 years ago

How many solutions does the nonlinear system of equations graphed below have?

Ostrovityanka [42]

Answer:

Option (A)

Step-by-step explanation:

Solution of two functions represented by the graph are the common points or point of intersection of the graphs.

From the graph attached,

Parabola and ellipse are intersecting each other at four points.

Therefore, solutions of the given non linear functions will be FOUR.

Option (A) will be the correct option.

4 0

3 years ago

Graph the following equation: y = 5

Irina-Kira [14]

If you graphed the equation: y = 5

You would put a straight line going horizontally through 5. Like this picture shows. If it were x = 5, for example, then it would go up and down through 5.

Let me know if this helped! Have a nice day!

6 0

3 years ago

Read 2 more answers

Carolyn leaves at 2:15 and drives 40 miles per hour. Marcel leaves at 2:45 and drives at 65 miles per hour. If they are travelin

Diano4ka-milaya [45]

Marcel leaves 30 minutes after Carolyn, leaving between them a gap of 40mph*0.5h=20miles
To catch up with the 20miles, Marcel uses the extra (65-40=25)mph that he has.
Therefore 20m/25mph=0.8h=0.8*60min=48minutes after 2:45, which is at 3:33, Marcel catches up to Carolyn.

8 0

3 years ago

According to data from a medical association, the rate of change in the number of hospital outpatient visits, in millions, in

GaryK [48]

Answer:

a) $f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+264,034,000$

b) f(t=2015) = 264,034,317.7

Step-by-step explanation:

The rate of change in the number of hospital outpatient visits, in millions, is given by:

$f'(t)=0.001155t(t-1980)^{0.5}$

a) To find the function f(t) you integrate f(t):

$\int \frac{df(t)}{dt}dt=f(t)=\int [0.001155t(t-1980)^{0.5}]dt$

To solve the integral you use:

$\int udv=uv-\int vdu\\\\u=t\\\\du=dt\\\\dv=(t-1980)^{1/2}dt\\\\v=\frac{2}{3}(t-1980)^{3/2}$

Next, you replace in the integral:

$\int t(t-1980)^{1/2}=t(\frac{2}{3}(t-1980)^{3/2})- \frac{2}{3}\int(t-1980)^{3/2}dt\\\\= \frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}+C$

Then, the function f(t) is:

$f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+C'$

The value of C' is deduced by the information of the exercise. For t=0 there were 264,034,000 outpatient visits.

Hence C' = 264,034,000

The function is:

$f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+264,034,000$

b) For t = 2015 you have:

$f(t=2015)=0.001155[\frac{2}{3}(2015)(2015-1980)^{1/2}-\frac{4}{15}(2015-1980)^{5/2}]+264,034,000\\\\f(t=2015)=264,034,317.7$

3 0

3 years ago