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2 years ago

C+12=15 idk this pls help

Mathematics

2 answers:

Rama09 [41]2 years ago

8 0

I think what you are looking for is what c equals and that would be c= 3 since 3 + 12 = 15

Vedmedyk [2.9K]2 years ago

4 0

Answer:

C = 3

Step-by-step explanation:

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The graph shows the cost of some taxi journeys<br><br> work out a formula for c in terms of n

Vitek1552 [10]

Answer:

$c=\frac{3}{5}(n)+2.5$

Step-by-step explanation:

see the attached figure to better understand the problem

Let

c -----> the cost of some taxi journeys

n ----> the number of miles

we know that

The graph is a line

The equation of a line in slope intercept form is

$c=m(n)+b$

where

m is the slope

b is the c-intercept (value of c when the value of n is equal to zero)

Observing the graph we have the points

(0,2.5), (5,5.5)

<em>Find the slope</em>

The formula to calculate the slope between two points is equal to

$m=\frac{c2-c1}{n2-n1}$

substitute the values

$m=\frac{5.5-2.5}{5-0}$

$m=\frac{3}{5}$

we have

$b=2.5$ -----> because the point (0,2.5) is the c-intercept

substitute in the equation of slope-intercept form

$c=\frac{3}{5}(n)+2.5$

4 0

3 years ago

Please solve it for me!! The answer is 97.3cm

Arada [10]

Answer:

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5 0

2 years ago

Read 2 more answers

Which of the following functions are solutions of the differential equation y'' + y = 3 sin x? (select all that apply)

DiKsa [7]

Answer:

Step-by-step explanation:

We must compute the derivatives and check if the equation is satisfied.

A. $y'=(3\sin x)'=3\cos x$ . Differentiate again to get $y''=(3\cos x)'=-3\sin x$ , then $y''+y=-3\sin x+3\sin x=0\neq 3\sin x$ so this choice of y doesn't solve the equation.

B. $y'=3\sin x+3x\cos x-5\cosx +5x\sin x=(5x+3)\sin x+(3x-5)\cos x$ and $y''=5\sin x+(5x+3)\cos x+3\cos x-(3x-5)\sin x=(10-3x)\sin x+(5x+6)\cos x$ , then $y''+y=10\sin x+6\cos x\neq 3\sin x$ so y is not a solution

C. $y'=\frac{-3}{2}\cos x+\frac{3}{2}x\sin x$ hence $y''=\frac{3}{2}\sin x+\frac{3}{2}\sin x+\frac{3}{2}x\cos x=3\sin x+\frac{3}{2}x\cos x$ . Then $y''+y=3\sin x+\frac{3}{2}x\cos x-\frac{3}{2}x\cos x=3\sin x$ so y is a solution.

D. $y'=-3\sin x$ and $y''=-3\cos x$ , then $y''+y=0$ thus y isn't a solution

E. $y'=\frac{3}{2}\sin x+\frac{3}{2}x\cos x$ hence $y''=\frac{3}{2}\cos x+\frac{3}{2}\cos x-\frac{3}{2}x\sin x=3\cos x-\frac{3}{2}x\cos x$ . Then $y''+y=3\cos x-\frac{3}{2}x\cos x-\frac{3}{2}x\cos x=3\cos x\neq 3\sin x$ then y is not a solution.