To find how many cm are in a ft, you have to use the method of unit rate:

so we know that 12 in. equals a ft. so,

1 ft
------
12 in.

and 2.25 cm equals an in. so,

1 in.
-------
2.54cm

then we put the rates together and get:

1ft 1 in.
---- X -------................we cancel out the in. units and get:
12 in. 2.54 cm

Solve and get 1 ft = 30.48 cm

4 0

3 years ago

What’s 0.17 repeating as a fraction

Murljashka [212]

If it's just the 7 that's repeating, let

x = 0.1777...

Then

10x = 1.777...

100x = 17.777...

100x - 10x = 17.777... - 1.777...

90x = 16

x = 16/90 = 8/45

If both 1 and 7 are repeating, let

x = 0.171717...

Then

100x = 17.171717...

100x - x = 17.171717... - 0.171717...

99x = 17

x = 17/99

7 0

2 years ago

Ice is placed around a bowl of water to lower the temperature. The equation D=−75t+22 D 7 5 t 22 shows the time, t, measured in

Minchanka [31]

Answer:

Step-by-step explanation:

D=-75t+22

8 0

3 years ago

If alpha and beta are the angles in the first quadrant tan alpha = 1/7 and sin beta =1/ root 10 then usind the formula sin (A +B

zysi [14]

Answer:

$\arcsin\left(\frac{129\sqrt{2}}{250}\right)\approx 0.8179$

Step-by-step explanation:

$\alpha \text{ and } \beta \text{ in Quadrant I}$

$\tan(\alpha)=\frac{1}{7} \text{ and } \sin(\beta)=\frac{1}{\sqrt{10}}=\frac{\sqrt{10} }{10}$

Using Pythagorean Identities:

$\boxed{\sin^2(\theta)+\cos^2(\theta)=1} \text{ and } \boxed{1+\tan^2(\theta)=\sec^2(\theta)}$

$\left(\frac{\sqrt{10} }{10} \right)^2+\cos^2(\beta)=1 \Longrightarrow \cos(\beta)=\sqrt{1-\frac{10}{100}} =\sqrt{\frac{90}{100}}=\frac{3\sqrt{10}}{10}$

$\text{Note: } \cos(\beta) \text{ is positive because the angle is in the first qudrant}$

$1+\left(\frac{1 }{7} \right)^2=\frac{1}{\cos^2(\alpha)} \Longrightarrow 1+\frac{1}{49}=\frac{1}{\cos^2(\alpha)} \Longrightarrow \frac{50}{49} =\frac{1}{\cos^2(\alpha)}$

$\Longrightarrow \frac{49}{50}=\cos^2(\alpha) \Longrightarrow \cos(\alpha)=\sqrt{\frac{49}{50} } =\frac{7\sqrt{2}}{10}$

$\text{Now let's find }\sin(\alpha)$

$\sin^2(\alpha)+\left(\frac{7\sqrt{2} }{10}\right)^2=1 \Longrightarrow \sin^2(\alpha) +\frac{49}{50}=1 \Longrightarrow \sin(\alpha)=\sqrt{1-\frac{49}{50}} = \frac{\sqrt{2}}{10}$