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3 years ago

What is the slope of the line? A. 3/4 B. -3/4 C. -4/3 D. 4/3

Mathematics

2 answers:

Sedbober [7]3 years ago

8 0

Answer:

d, you go up for and over 3

pashok25 [27]3 years ago

4 0

(0,0) (3,4)
(4-0)/(3-0) = 4/3
The slope is 4/3

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Hexagon IJKLMN is shown on the coordinate plane below:

MissTica

<h2>Answer:</h2>

Option: A is the correct answer.

The ordered pair of point I' are:

A) (-0.8,2.4)

<h2>Step-by-step explanation:</h2>

We know than if any point A with vertices as (a,b) are dilated by a scale factor of k then the vertices of the transformed point is:

A(a,b) → A'(ka,kb)

We are given vertices of point I as:

I(-2,6)

Hence, the vertices of the transformed point I' after dilating by a scale factor of 2/5 is:

I(-2,6) → I'(-2×(2/5),6×(2/5))

⇒ I(-2,6) → I'(-0.8,2.4)

Hence, option: A is the answer.

A) (-0.8,2.4)

3 0

3 years ago

Read 2 more answers

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.

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