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S_A_V [24]
2 years ago
13

Two trains are traveling at a constant rate. Which train has the greater speed?

Mathematics
1 answer:
natali 33 [55]2 years ago
3 0
A point on the graph (for train b) that is exact is (2,250). From the origin, the rise is 250 and the run is 2. Rise/run would give us 250/2, which is 125. This means that train B travels 125mi every hour. For train A, we subtract the distance from hour 2 from hour 3 to get the miles per hour (because 3-2=1, subtracting the distance of hour 3 from 2 will give us the distance in 1 hour). 180-120= 60, meaning train A travels at 60mph. 60<125, meaning A
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= -2

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2 years ago
The equation for y=-2x+5 that is paralell passes through (1,-4)?
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Make r the subject of the formula:<br> U=v+2t
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U=v+ 2t=
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8 0
3 years ago
Find the maximum profit and the number of units that must be produced and sold in order to yield the maximum profit. Assume that
V125BC [204]

Answer:

-1.39

Step-by-step explanation:

Revenue and cost as a function of units sold are u(x) = 9x-2x^{2}andc(x)=x^{3}-3x^{2}+4x+1  respectively.

we are have to know for which value or input units are these functions at maximum which translates to for how many units is the revenue maximum and for how many same units is our cost minimum.

5 0
3 years ago
What is the result of substituting for y in the bottom equation<br> y=x+3<br> y=x^2+2x-4
8_murik_8 [283]

The solutions are (2.1925, 5.1925) and (-3.1925, -0.1925)

<em><u>Solution:</u></em>

Given that,

y = x + 3 ------- eqn 1\\\\y = x^2 + 2x - 4  ----- eqn 2

<em><u>We have to substitute eqn 1 in eqn 2</u></em>

x + 3 = x^2 + 2x - 4

\mathrm{Switch\:sides}\\\\x^2+2x-4=x+3\\\\\mathrm{Subtract\:}3\mathrm{\:from\:both\:sides}\\\\x^2+2x-4-3=x+3-3\\\\\mathrm{Simplify}\\\\x^2+2x-7=x\\\\\mathrm{Subtract\:}x\mathrm{\:from\:both\:sides}\\\\x^2+2x-7-x=x-x\\\\\mathrm{Simplify}\\\\x^2+x-7=0

\mathrm{Solve\:with\:the\:quadratic\:formula}\\\\\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\\\x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\\mathrm{For\:}\quad a=1,\:b=1,\:c=-7\\\\x =\frac{-1\pm \sqrt{1^2-4\cdot \:1\left(-7\right)}}{2\cdot \:1}

x = \frac{-1 \pm \sqrt{ 1 + 28}}{2}\\\\x = \frac{ -1 \pm \sqrt{29}}{2}

x = \frac{ -1 \pm 5.385 }{2}\\\\We\ have\ two\ solutions\\\\x = \frac{ -1 + 5.385 }{2}\\\\x = 2.1925

Also\\\\x = \frac{ -1 - 5.385 }{2}\\\\x = -3.1925

Substitute x = 2.1925 in eqn 1

y = 2.1925 + 3

y = 5.1925

Substitute x = -3.1925 in eqn 1

y = -3.1925 + 3

y = -0.1925

Thus the solutions are (2.1925, 5.1925) and (-3.1925, -0.1925)

6 0
3 years ago
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