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3 years ago

Maths Mate Help!

Mathematics

1 answer:

grigory [225]3 years ago

3 0

Answer:

Step-by-step explanation:

Rainforests = 6%

Other habitats = 100% - 6% = 94%

The ratio would there fore be:

6%/94%

In fraction:

6% = 6/100 = 3/50

94% = 94/100 = 47/50

So,

$\frac{\frac{3}{50}}{\frac{47}{50}}=\frac{3}{50}*\frac{50}{47}=\frac{3}{47}$

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Which step is incorrect when solving the equation 2(x+9)=4(x+7)+2?

iragen [17]

Answer:

3. 2x+18=4x+26

Step-by-step explanation:

2(x+9)=4(x+7)+2

2x+18=4x+28+2

2x+18=4x+30

The step that made an error in solving this linear equation is step 3.

Hope this helps ;) ❤❤❤

8 0

4 years ago

George has a bag with 4 mint sticks, 3 jelly treats, and 13 fruit tart chews. If he eats one piece every 5 minutes, what is the

slega [8]

Answer:

12/625 or 1.9%

Step-by-step explanation:

Adding all of the candy pieces together, you get 25, the denominator.

3/25*4/25=12/625

12/625=.0192

.0192*100=1.92%

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4 years ago

The value of the 8 in 8.4 is 100 times greater than the 8 in

anastassius [24]

The answer is= 0.08

I might be wrong but I think it is correct! :)

Have a great day!

5 0

4 years ago

Read 2 more answers

Solve -2 1/3 - (-5) ANSWER PLZ

Jlenok [28]

Answer:

Step-by-step explanation:

2.6666666667

6 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