Help on these plzzzzz it is really hard

Find the reference angle of <br><img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B%20-%20%5Cpi%7D%7B7%7D%20" id="TexFormula1" title

Semenov [28]

Answer:

π7

Step-by-step explanation:

since is in the first quadrant, the reference angle is π7

3 0

3 years ago

Create a real-life situation that can be describe with the unit rate 9:1. Describe how quantities are related.

iris [78.8K]

I have 81¥ and I'm going to the US, so I need to convert my Yen into US Dollars, if i get 1 USD (US Dollar) for every 9¥, how many USD will i have if i convert all of my Yen?

4 0

3 years ago

A piece if wire 40cm is bent to form a right angled triangle whose hypotenuse is 17cm long find the other two sides of triangle

hjlf

Answer: 150mm

Step-by-step explanation:

Take the two adjecent sides of the triangle to be x and y and X is the angle opposite the side, x.

From the trigonometric identities for a right-angled triangle, it is known that

x=170sin X and;

y=170cos X

Thus the perimeter of the triangle is given by

170sin X + 170cos X + 170 = 400

170(sin X + cos X) = 400 - 170

170(sin X + cos X) = 230

sin X + cos X = 230/170 = 1.3529

Squaring both sides hence,

sin^2(X) + 2sin X.cos X + cos^2(X) = 1.8304

Recall that sin^2(X) + cos^2(X) = 1 and 2sin X.cos X = sin 2X.

Thus, the former expression gives;

1 + sin 2X = 1.8304

sin 2X = 0.8304

2X = arcsin (0.8304)

2X = 56.14° or (180 - 56.14)° = 56.14° or 123.86° for X existing within a triangle.

X = 56.14/2 or 123.86/2

X = 28.07° or 61.93°

Thus the two adjacent sides are ;

x = 170 sin (28.07) = 80mm

y = 170 cos (28.07) = 150mm

Using the value of 61.93° as X would give the same values but interchanged for x and y.

I hope you find this useful.

5 0

3 years ago

Which situation can be repsented by the inequailty?

Georgia [21]

I think its either A or D

Hope this helps!

4 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

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