Ask question

Ask question

All categories

1 year ago

10

(05.05 HC) Figure ABCD has vertices A(-3, 2), B(2, 2), C(2, -4), and D(-3, -2). What is the area of Figure ABCD?

1 answer:

irga5000 [103]1 year ago

6 0

Answer:

I don't know how to answer this, sorry!

Step-by-step explanation:

.

You might be interested in

6. Order these fractions from least to<br> greatest:<br> 2/3<br> 7/12<br> 3/4

Oksi-84 [34.3K]

2/3 = 8/12

7/12 = 7/12

3/4 = 9/12

7/12 < 8/12 < 9/12

So, 7/12 < 2/3 < 3/4

8 0

2 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

jacob has a box of 2 hundred cubes he also has a bag of 40 cubes. how many trains of 10 cubes can hw make?

Verizon [17]

200+40=240
each train cost 10
take 240/10=24trains

6 0

3 years ago

HELP PLEASEEEEE!!!! ILL GIVE BRAINLIEST

MAVERICK [17]

Answer:

they cost 13.86666666666667

Step-by-step explanation:

6 0

3 years ago

The perimeter of a rectangular swimming pool is 414 meters. The length of the pool is 7 meters more than four times the width. F

PolarNik [594]

Answer:

Width = 40 meters

Length = 167 meters

Step-by-step explanation:

Let

Width of the pool = x meters

Then

Length of the pool = 4x + 7 meters

and

Perimeter $=2(x+4x+7)=2(5x+7)=10x+14$ meters

Hence,

$10x+14=414\\ \\10x=414-14\\ \\10x=400\\ \\x=40\ meters\\ \\4x+7=4\cdot 40+7=167\ meters$

5 0

2 years ago