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3 years ago

Simplify 1/4 (1-2/3)^2+1/3

Mathematics

2 answers:

tiny-mole [99]3 years ago

7 0

In exact form: 13/36

in decimal form: 0.36111111...

muminat3 years ago

5 0

$\frac{1}{4} * (1-2/3)^{2} + \frac{1}{3}$

= $\frac{1}{4} * ( 1 - \frac{4}{3} + \frac{4}{9} ) + \frac{1}{3}$

= $\frac{1}{4} - \frac{1}{3} + \frac{1}{9} + \frac{1}{3}$

= $\frac{1}{4} + \frac{1}{9}$

= $\frac{9 + 4}{9 * 4}$

= $\frac{13}{36}$

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3 years ago

A line through the origin and (10,4) is showin in the standard (x,y) coordinate plane below. The acute angle between the line an

Snowcat [4.5K]

Answer:

$\cos\theta=\frac{10}{\sqrt{116}}$

Step-by-step explanation:

To solve this, we can imagine a line drawn directly down from the point (10,4) to the x-axis. This would make a right triangle with length 10 and height 4.

$\cos\theta$ is equal to the length of adjacent leg to the angle divided by the length of the hypotenuse of the triangle. From the picture, we can see the adjacent leg is 10.

We can use Pythagorean's Theorem to find the hypotenuse (let $c$ represent the length of the hypotenuse):

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$c=\sqrt{116}$

Thus, we know the adjacent is 10, and the hypotenuse is $\sqrt{116}$ . This means $\cos\theta=\frac{10}{\sqrt{116}}$ .

Hope this helps!

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2 years ago

Jerry walked a dog from 6:40 am to 7:30 pm one day. If he was paid at the rate of 6$ per hour, how much did he earn that day?

Tom [10]

Answer:

Explanation:

Since Jerry walked the dog from 6:40am to 7:30am, the total minutes that he walked the dog was 50 minutes. Also a note, an hour is 60 minutes.

Since he is paid at a rate of $6 per 60 minutes, we can make a proportion.

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From here, you have to solve for x. So cross multiply, and then simplify.

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60x60=50⋅6/60

x=5⋅6/6

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3 years ago

Read 2 more answers

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Answer:

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Step-by-step explanation:

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3 years ago

Solve the triangle A = 2 B = 9 C =8

VARVARA [1.3K]

Answer:

$\begin{gathered} A=\text{ 12}\degree \\ B=\text{ 114}\degree \\ C=54\degree \end{gathered}$

Step-by-step explanation:

To calculate the angles of the given triangle, we can use the law of cosines:

$\begin{gathered} \cos (C)=\frac{a^2+b^2-c^2}{2ab} \\ \cos (A)=\frac{b^2+c^2-a^2}{2bc} \\ \cos (B)=\frac{c^2+a^2-b^2}{2ca} \end{gathered}$

Then, given the sides a=2, b=9, and c=8.

$\begin{gathered} \cos (A)=\frac{9^2+8^2-2^2}{2\cdot9\cdot8} \\ \cos (A)=\frac{141}{144} \\ A=\cos ^{-1}(\frac{141}{144}) \\ A=11.7 \\ \text{ Rounding to the nearest degree:} \\ A=12º \end{gathered}$

For B:

$\begin{gathered} \cos (B)=\frac{8^2+2^2-9^2}{2\cdot8\cdot2} \\ \cos (B)=\frac{13}{32} \\ B=\cos ^{-1}(\frac{13}{32}) \\ B=113.9\degree \\ \text{Rounding:} \\ B=114\degree \end{gathered}$ $\begin{gathered} \cos (C)=\frac{2^2+9^2-8^2}{2\cdot2\cdot9} \\ \cos (C)=\frac{21}{36} \\ C=\cos ^{-1}(\frac{21}{36}) \\ C=54.3 \\ \text{Rounding:} \\ C=\text{ 54}\degree \end{gathered}$

3 0

1 year ago