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Natali5045456 [20]

3 years ago

6

Can you help me with this? I don't seem to understand..

1 answer:

gulaghasi [49]3 years ago

3 0

All you have to do is set up 2 proportions, where one paint you know and the other you do not know and make it equal to the ratio as 2 blue : 5 yellow

For example: 1 blue/x yellow = 2 blue/5 yellow, and then cross multiply and solve for x.

Do this for all unknowns, blue and yellow paint making it equal to the ratio and solve for the unknown.

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The cookie recipe calls for 1/2 cup of flour to make 8 servings. How many cups of flour are needed to make 20 servings of cookie

Musya8 [376]

20/8=2.5 1/2+1/2+1/2=1 1/2

5 0

3 years ago

Read 2 more answers

Miguel deposits $2000 into a bank account that earns simple interest. He writes the expression 2000 + 2000(0.015)(t/12) to repre

oee [108]

Answer:

based on the equation he would make $2.5 in the first month but realistically he would earn around $30 in the first month

Step-by-step explanation:

2000+2000(.015)(1/12)=2002.5

2002.5-2000=2.5

8 0

3 years ago

The liquid base of an ice cream has an initial temperature of 86°C before it is placed in a freezer with a constant temperature

Karolina [17]

The temperature of the ice cream 2 hours after it was placed in the freezer is 37.40 °C

From Newton's law of cooling, we have that

$T_{(t)}= T_{s}+(T_{0} - T_{s})e^{kt}$

Where

$(t) = \ time$

$T_{(t)} = \ the \ temperature \ of \ the \ body \ at \ time \ (t)$

$T_{s} = Surrounding \ temperature$

$T_{0} = Initial \ temperature \ of \ the \ body$

$k = constant$

From the question,

$T_{0} = 86 ^{o}C$

$T_{s} = -20 ^{o}C$

∴ $T_{0} - T_{s} = 86^{o}C - -20^{o}C = 86^{o}C +20^{o}C$

$T_{0} - T_{s} = 106^{o} C$

Therefore, the equation $T_{(t)}= T_{s}+(T_{0} - T_{s})e^{kt}$ becomes

$T_{(t)}=-20+106 e^{kt}$

Also, from the question

After 1 hour, the temperature of the ice-cream base has decreased to 58°C.

That is,

At time $t = 1 \ hour$ , $T_{(t)} = 58^{o}C$

Then, we can write that

$T_{(1)}=58 = -20+106 e^{k(1)}$

Then, we get

$58 = -20+106 e^{k(1)}$

Now, solve for $k$

First collect like terms

$58 +20 = 106 e^{k}$

$78 =106 e^{k}$

Then,

$e^{k} = \frac{78}{106}$

$e^{k} = 0.735849$

Now, take the natural log of both sides

$ln(e^{k}) =ln( 0.735849)$

$k = -0.30673$

This is the value of the constant $k$

Now, for the temperature of the ice cream 2 hours after it was placed in the freezer, that is, at $t = 2 \ hours$

From

$T_{(t)}=-20+106 e^{kt}$

Then

$T_{(2)}=-20+106 e^{(-0.30673 \times 2)}$

$T_{(2)}=-20+106 e^{-0.61346}$

$T_{(2)}=-20+106\times 0.5414741237$

$T_{(2)}=-20+57.396257$

$T_{(2)}=37.396257 \ ^{o}C$

$T_{(2)} \approxeq 37.40 \ ^{o}C$

Hence, the temperature of the ice cream 2 hours after it was placed in the freezer is 37.40 °C

Learn more here: brainly.com/question/11689670

6 0

2 years ago

Read 2 more answers

A system whose graphs have the same slope but different y- intercepts

almond37 [142]

X intersept form........

8 0

3 years ago

A bag contains 12 red marbles The ratio of red marbles to blue marbles is 4:3?

Ostrovityanka [42]

There are 9 blue marbles.

Step-by-step explanation:

12÷4=3

3×3=9

8 0

2 years ago