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3 years ago

What is the slope in the equations: y = 4x + 3

Mathematics

2 answers:

mylen [45]3 years ago

5 0

Answer:

Step-by-step explanation:

y = mx + b

m represents the slope, and as we can see in the equation, the number 4 is there instead of m, so therefore 4 is the slope

Anestetic [448]3 years ago

4 0

Using the slope-intercept form, the slope is 4

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In order to circumscribe a circle on a triangle, which line must you construct?

Viktor [21]

The answer is the option d.Perpendicular bisector.

Explanation:

By definition, a circumscribe circle passes through the three vertices of the triangle and to its constrution, you have to draw the triangle and then construct a perpendicular bisector to each side of it. The point where these perpendicular bisectors intersect each other, will be the center of the circle. Then, you should place the compass on the center point to draw the circle.

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3 years ago

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At the beginning of the day, it was -14 degrees. Throughout the day, the temperature increased 16 degrees. What was the temperat

konstantin123 [22]

Answer:

The temperature at the end of the day is 2 degrees

Step-by-step explanation:

Here, we are interested in calculating the temperature at the end of the day.

In the question, we are told that at the beginning of the day, temperature was -14 degrees.

But throughout the day, the temperature increased by 16 degrees;

Thus, the temperature at the end of the day = -14 degrees + 16 degrees = 2 degrees

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3 years ago

A cake is removed from a 310°F oven and placed on a cooling rack in a 72°F room. After 30 minutes the cake's temperature is 220°

Fynjy0 [20]

Answer:

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.

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