All categories

3 years ago

Ill give 10 points if you answer

Mathematics

1 answer:

scoundrel [369]3 years ago

5 0

Hi there!

The answer is x = 6. All you have to do is subtract 83 from 127 because 127 is the total of the two parts of the equation. Now using the expression they give you - since you already know what the missing angle degree is for JKM- you just plug it in. 9x-10=44.

Add 10 to both sides to cancel out the ten.

You then have 9x = 54.

Divide both sides by 9.

Leaving you with x = 6.

Hope this helps !

You might be interested in

Bring the help!!!!!!!!!!!!!!!!!

matrenka [14]

The answer to the problem is -8.58.

8 0

3 years ago

18 POINTS will rate brainiest 6th grade math

stira [4]

Answer:

18 ft^2

Step-by-step explanation:

The area of a parallelogram is A=bh.

The area would be A = 9*4.

A = 36

To find the area of one trapezoid you divide the area of the parallelogram by 2.

36/2 = 18 ft^2

5 0

3 years ago

2<3p-7< 11 what would be the correct answer for this qustion

Leni [432]

Answer:

Hey! Heres your answer -

Step-by-step explanation:

I hope this helps you! :)

3 0

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.

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

4 0

3 years ago

What is the remainder when f(x) = x2 + 14x − 8 is divided by (x − 3)?

ser-zykov [4K]

X^2 +14x -8 / x-3

= x + 17 with a remainder of 43

See attached picture for solution.

3 0

3 years ago