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tekilochka [14]
2 years ago
8

Write an equation in point-slope form that has a y-intercept at (0, 12) and also contains the point (1,2)

Mathematics
1 answer:
harkovskaia [24]2 years ago
6 0

Answer:

y-2=-10(x-1).

Step-by-step explanation:

The point slope form is y-y_1 = m(x-x_1). Now we find the slope, which is 2-12 (-10)/1-0(1). This gives us -10. Now we plug the coordinates from one of the points and the slope to get y-2=-10(x-1).

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Energy released as a gas changes into a liquid is called what? (#9)
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Energy will be released to change from liquid to solid (fusion), gas to liquid (condensation), or gas to solid. Latent heat of evaporation is the energy used to change liquid to vapor. IMPORTANT: The temperature does not change during this process, so heat added goes directly into changing the state of the substance

Step-by-step explanation:

4 0
3 years ago
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Jason tossed a fair coin 3 times. what is the probability of getting a head and two tails in any order?
mixer [17]

Step-by-step explanation:

fair coin is two sided Head H and Tail T

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4 0
2 years ago
If a plane can travel 480 miles per hour with the wind and 400 miles per hour against the wind find the speed of the wind and th
pychu [463]

he is going 80 mph because he can go 480 mph and the wind is against him 400 mph and 480-400=80

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7 0
3 years ago
The volume of a particular die is 12100 mm3. Use the fact that 10 mm equals 1 cm to convert this volume to cm3.
igor_vitrenko [27]
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10 kiloliter to liters, the result is 10000 liters

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8 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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