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2 years ago

12 POINTS! WILL MARK BRAINIEST! IF YOU TAKE THE POINTS I WILL REPORT PLEASE ASAP! QUESTION IN PIC

Mathematics

1 answer:

tamaranim1 [39]2 years ago

3 0

6-9y. The fourth choice

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Find the formula for the area of the shaded region in the figure.<br> x² = 4py<br> ہے

Alexandra [31]

The formula for the area of the shaded region in the figure.

x² = 4py is mathematically given as

$A=\frac{8}{3} \sqrt{p h} \cdot h$

<h3>What is the formula for the area of the shaded region in the figure?</h3>

Parameters

$x^{2}=4 p y$

$y=h$

$\therefore x^{2}=4 p h \Rightarrow x=2 \sqrt{p h} \\$

Generally, the equation for the Area is mathematically given as

$& A=2 \int_{0}^{2 \sqrt{p h}}\left(h-\frac{x^{2}}{4 p}\right) d x \\\\& A=2\left(h x-\frac{x^{3}}{12 p}\right]_{0}^{2 \sqrt{p h}} \\\\& A=2\left[\left[h 2 \sqrt{p h}-\frac{1}{12 p} \cdot(2 \sqrt{p h})^{3}-0\right]\right. \\$

$&A=2\left[2 h^{3 / 2} \sqrt{p}-\frac{8(\sqrt{p})^{3}(\sqrt{h})^{3}}{12 p}-0\right] \\\\&A=2\left[2 h^{k / 2} \sqrt{p}-\frac{2}{3} \sqrt{p} \cdot h^{3 / 2}\right] \\\\&A=2 \times \frac{4 \sqrt{p} \cdot h^{3 / 2}}{3} \\\\&A=\frac{8}{3} \cdot \sqrt{p} \cdot \sqrt{h} \cdot h \\\\&A=\frac{8}{3} \sqrt{p h} \cdot h$

$A=\frac{8}{3} \sqrt{p h} \cdot h$

In conclusion, the equation for the Area is

$A=\frac{8}{3} \sqrt{p h} \cdot h$

Read more about Area

brainly.com/question/27683633

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5 0

1 year ago

Triangle A is the pre-image, and triangle B is the image. Find the scale factor for this dilation if the center of dilation is a

BigorU [14]

Answer:

The scale factor is 3

Step-by-step explanation:

6/2=3 (one of the triangles sides)

8 0

2 years ago

What is the algebraic way to solve -10(x-17) = -3 ? Please provide the correct solution :)

suter [353]

Answer:173/10

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

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

Please help!!!!<br><br> Solve the formula A = /w for l.

alexdok [17]

A = wl

Simply divide both sides by w. This is done to isolate the variable we are solving for.

A/w = (wl)/w

A/w = l

Answer: Choice C

6 0

3 years ago