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liraira [26]
2 years ago
10

What are the solutions to the system of equations? y = 2x² + 6x - 10 \ y = y = -x + 5

Mathematics
1 answer:
Akimi4 [234]2 years ago
3 0

Answer:

x= 1.5, y= 3.5

or x= -5, y= 10

Step-by-step explanation:

Let's solve by substitution.

Label the two equations:

y= 2x² +6x -10 -----(1)

y= -x +5 -----(2)

Substitute (2) into (1):

2x² +6x -10= -x +5

2x² +6x +x -10 -5= 0

Simplify:

2x² +7x -15= 0

Factorise:

(2x -3)(x +5)= 0

2x -3= 0 or x +5= 0

2x= 3 or x= -5

x= 1.5

Now that we have found the value of x, we can find the value of y through substitution.

Substitute into (2):

y= -1.5 +5 or y= -(-5) +5

y= 3.5 or y= 10

Supplementary:

Do check out the following should you wish to learn more about solving quadratic equations!

  • brainly.com/question/12187987

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The time is 135 min.

Step-by-step explanation:

For this situation we are going to use Newton's Law of Cooling.

Newton’s Law of Cooling states that the rate of temperature of the body is proportional to the difference between the temperature of the body and that of the surrounding medium and is given by

T(t)=C+(T_0-C)e^{kt}

where,

C = surrounding temp

T(t) = temp at any given time

t = time

T_0 = initial temp of the heated object

k = constant

From the information given we know that:

  • Initial temp of the cake is 310 °F.
  • The surrounding temp is 72 °F.
  • After 30 minutes the cake's temperature is 220 °F.

We want to find the time, in minutes, since the cake's removal from the oven, at which its temperature will be 100°F.

To do this, first, we need to find the value of k.

Using the information given,

220=72+(310-72)e^{k\cdot 30}\\\\72+238e^{k30}=220\\\\238e^{k30}=148\\\\e^{k30}=\frac{74}{119}\\\\\ln \left(e^{k\cdot \:30}\right)=\ln \left(\frac{74}{119}\right)\\\\k\cdot \:30=\ln \left(\frac{74}{119}\right)\\\\k=\frac{\ln \left(\frac{74}{119}\right)}{30}

T(t)=72+(310-72)e^{(\frac{\ln \left(\frac{74}{119}\right)}{30}\cdot t)}

Next, we find the time at which the cake's temperature will be 100°F.

100=72+(310-72)e^{(\frac{\ln \left(\frac{74}{119}\right)}{30}\cdot t)}\\72+238e^{\frac{\ln \left(\frac{74}{119}\right)}{30}t}=100\\238e^{\frac{\ln \left(\frac{74}{119}\right)}{30}t}=28\\e^{\frac{\ln \left(\frac{74}{119}\right)}{30}t}=\frac{2}{17}\\\ln \left(e^{\frac{\ln \left(\frac{74}{119}\right)}{30}t}\right)=\ln \left(\frac{2}{17}\right)\\\frac{\ln \left(\frac{74}{119}\right)}{30}t=\ln \left(\frac{2}{17}\right)\\t=\frac{30\ln \left(\frac{2}{17}\right)}{\ln \left(\frac{74}{119}\right)}\approx 135.1

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3 years ago
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