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3 years ago

How do i convert a fraction into a repeating decimal???

Mathematics

1 answer:

Mariulka [41]3 years ago

5 0

You simply divide the numerator by the denominator. Based on the numbers, you’ll get a decimal, repeating or otherwise

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What is 34 divided by 45

Mrrafil [7]

34 ÷ 45 $\frac{34}{45}$ = 0.7555555555556 ≈ 0.76

7 0

3 years ago

What is the solution?

AfilCa [17]

Answer:

Step-by-step explanation:

4 0

3 years ago

OC and OR are similar. Find the length of QP.

Elena L [17]

Given:

Circle C and circle R are similar.

The length of arc AB is $s = \frac{22 \pi}{9}$

The radius of circle C (AC) = 4 unit

The radius of circle R (QR) =6 unit

To find the length of arc QP.

Formula

The relation between s, r and $\theta$ is

$arclength = 2\pi r \frac{\theta}{360}$

where,

s be the length of the arc

r be the radius

$\theta$ be the angle.

Now,

For circle C

Taking r = 4

According to the problem,

$2 \pi r \frac{\theta}{360} = \frac{22 \pi}{9}$

or, $2r \frac{\theta}{360} = \frac{22}{9}$ [ eliminating $\theta$ from both side]

or, $\theta = \frac{(22)(360)}{(9)(2)(4)}$

or, $\theta = 110^\circ$

Again,

For circle R

Taking, r = 6 and $\theta = 110^\circ$ we get,

The length of arc QP is

$arc length = 2\pi (6)(\frac{110}{360} )$

or, $arclength = \frac{11 \pi}{3}$

Hence,

The length of QP is $\frac{11 \pi}{3}$ . Option C.

5 0

3 years ago

Is it aaaaaaaa or dddddddd

nika2105 [10]

Its none
because a triangle makes up 180 degrees hope this helps and im not a dumby

4 0

3 years ago

Read 2 more answers

Quadrant:

NISA [10]

<h2>✒️Area Between Curves</h2>

$\small\begin{array}{ |c|c} \hline \bold{Area\ Between\ Curves} \\ \\ \textsf{Solving for the intersection of }\rm y = x^2 + 2\textsf{ and }\\ \rm y = 4, \\ \\ \qquad \begin{aligned} \rm y_1 &=\rm y_2 \\ \rm x^2 + 2 &=\rm 4 \\ \rm x^2 &= \rm 2 \\ \rm x &=\rm \pm \sqrt{2} \end{aligned} \\ \\ \textsf{We only need the first quadrant area bounded} \\ \textsf{by the given curves so the integral for the area} \\ \textsf{would then be} \\ \\ \boldsymbol{\displaystyle \rm A = \int_{\ a}^{\ b} {\left( \begin{array}{c}\text{upper} \\ \text{function}\end{array} \right) - \left( \begin{array}{c} \text{lower} \\ \text{function} \end{array} \right)\ dx}} \\ \\ \displaystyle \rm A = \int_{0}^{\sqrt{2}} \Big[4 - (x^2 + 2)\Big]\ dx \\ \\ \displaystyle \rm A = \int_{0}^{\sqrt{2}} (2 - x^2)\ dx \\ \\ \rm A = \left[2x - \dfrac{x^3}{3}\right]_{0}^{\sqrt{2}} \\ \\ \rm A = 2\sqrt{2} - \dfrac{\big(\sqrt{2}\big)^3}{3} \\ \\ \rm A = 2\sqrt{2} - \dfrac{2\sqrt{2}}{3} \\ \\\red{\boxed{\begin{array}{c} \rm A = \dfrac{4\sqrt{2}}{3}\textsf{ sq. units} \\ \textsf{or} \\ \rm A \approx 1.8856\textsf{ sq. units} \end{array}}} \\\\\hline\end{array}$

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5 0

2 years ago