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liq [111]
3 years ago
5

Jeremy walked 6/8 of the way to school and ran the rest of the way. What fraction in simplest form shows the part of the way tha

t Jeremy walked?
Mathematics
1 answer:
Elan Coil [88]3 years ago
5 0
To simplest form of a fraction you find the lowest common factor, usually I use this trick: if both numbers are even I keep dividing by 2 to the numerator and denominator till I can't anymore
In this case:
6 divided by 2= 3
---                     ---
8 divided by 2= 4

now I can't simplify an further because 3 isn't an even number, therefore its the simplest form
remember= my trick only works if both number are even, if not you'll have to find the lowest common factor and divide them by that
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If A ABC and AABD are equilateral, which line segment is a 60° clockwise rotation of AB about the
Svet_ta [14]

Answer:

I think it is AC

Step-by-step explanation:

5 0
3 years ago
Miranda needs to add 68 + 19 mentally but is struggling. If she knows that 68 + 20 = 88, how can she use this to help?
stealth61 [152]
She can use 68 + 20 = 88 to help by noticing that the 20 is ahead of 19 by 1. She can easily mentally turn the 20 into a 19 and realized that the 88 would become 87.
7 0
2 years ago
If sin θ = 1 over 4 and tan θ > 0, what is the value of cos θ? (1 point)
Dima020 [189]

Answer:

There are two options: b) \frac{\sqrt{15}}{4}, c) \sqrt{15}.

Step-by-step explanation:

By Trigonometry, we know that function tangent is equal to the following identity in terms of functions sine and cosine:

\tan \theta = \frac{\sin \theta}{\cos \theta} (1)

If we know that \tan \theta > 0 and \sin \theta = \frac{1}{4}, then \cos \theta > 0. Hence, we have two possibilities: b) \frac{\sqrt{15}}{4}, c) \sqrt{15}.

4 0
3 years ago
Camila wanted to create a second garden to grow even more vegetables. To determine the dimensions of her second garden, she used
OlgaM077 [116]

Using proportions, considering original measures of 10 feet, we have that:

  • The horizontal measurement of the second garden is of 7.5 feet.
  • The vertical measurement of the second garden is of 15 feet.

<h3>What is a proportion?</h3>

A proportion is a fraction of total amount.

In this problem, the total amount is the original measures of 10 feet.

The horizontal measurement has a scale factor of 75%, hence:

h = 0.75(10) = 7.5

The vertical measurement has a scale factor of 150%, hence:

h = 1.5(10) = 15

You can learn more about proportions at brainly.com/question/24372153

3 0
2 years ago
Suppose integral [4th root(1/cos^2x - 1)]/sin(2x) dx = A<br>What is the value of the A^2?<br><br>​
Alla [95]

\large \mathbb{PROBLEM:}

\begin{array}{l} \textsf{Suppose }\displaystyle \sf \int \dfrac{\sqrt[4]{\frac{1}{\cos^2 x} - 1}}{\sin 2x}\ dx = A \\ \\ \textsf{What is the value of }\sf A^2? \end{array}

\large \mathbb{SOLUTION:}

\!\!\small \begin{array}{l} \displaystyle \sf A = \int \dfrac{\sqrt[4]{\frac{1}{\cos^2 x} - 1}}{\sin 2x}\ dx \\ \\ \textsf{Simplifying} \\ \\ \displaystyle \sf A = \int \dfrac{\sqrt[4]{\sec^2 x - 1}}{\sin 2x}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\sqrt[4]{\tan^2 x}}{\sin 2x}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\sqrt{\tan x}}{\sin 2x}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\sqrt{\tan x}}{\sin 2x}\cdot \dfrac{\sqrt{\tan x}}{\sqrt{\tan x}}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\tan x}{\sin 2x\ \sqrt{\tan x}}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\dfrac{\sin x}{\cos x}}{2\sin x \cos x \sqrt{\tan x}}\ dx\:\:\because {\scriptsize \begin{cases}\:\sf \tan x = \frac{\sin x}{\cos x} \\ \: \sf \sin 2x = 2\sin x \cos x \end{cases}} \\ \\ \displaystyle \sf A = \int \dfrac{\dfrac{1}{\cos^2 x}}{2\sqrt{\tan x}}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\sec^2 x}{2\sqrt{\tan x}}\ dx, \quad\begin{aligned}\sf let\ u &=\sf \tan x \\ \sf du &=\sf \sec^2 x\ dx \end{aligned} \\ \\ \textsf{The integral becomes} \\ \\ \displaystyle \sf A = \dfrac{1}{2}\int \dfrac{du}{\sqrt{u}} \\ \\ \sf A= \dfrac{1}{2}\cdot \dfrac{u^{-\frac{1}{2} + 1}}{-\frac{1}{2} + 1} + C = \sqrt{u} + C \\ \\ \sf A = \sqrt{\tan x} + C\ or\ \sqrt{|\tan x|} + C\textsf{ for restricted} \\ \qquad\qquad\qquad\qquad\qquad\qquad\quad \textsf{values of x} \\ \\ \therefore \boxed{\sf A^2 = (\sqrt{|\tan x|} + c)^2} \end{array}

\boxed{ \tt   \red{C}arry  \: \red{ O}n \:  \red{L}earning}  \:  \underline{\tt{5/13/22}}

4 0
2 years ago
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