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3 years ago

7

For every pound a company spends on advertising, it spends £0.37 on its website. Express the amount spent on advertising to its

website as a ratio in its simplest form.

1 answer:

denis23 [38]3 years ago

3 0

it would either be 100:37 or it would be 37: 100. i'm not sure which way.

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Find the distance between these points: (-3, 8) and (-5,1)

irina1246 [14]

Answer:

7.28011

Step-by-step explanation:

X1 =-3

X2 =-5

Y1 =8

Y2 =1

Distance Formula = √(X2-X1)²+ (Y2-y1)²

Step 1. calculate (X2-X1)²

(-5 - -3)2 = 4

Step 2. calculate (Y2-Y1)²

(1 - 8)2 = 49

Step 3. calculate √4 + 49

= 7.28011

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3 years ago

The given lengths are two sides of a right triangle. All three side lengths of the triangle are integers, and together they form

mel-nik [20]

The correct answer is “A” 40,leg

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3 years ago

Lol can someone help asap i need this done for a quiz, i’ll give brainliest.

maksim [4K]

Answer:

8.06

Step-by-step explanation:

$\sqrt{4^{2}+7^{2} }$

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3 years ago

The range, a measure of variability, Group of answer choices is the difference between the largest (L) and smallest (S) value in

Vlad [161]

<u>The correct answer is </u><u>all of the above.</u>

Why is the range the most convenient measure of variability?

Your data's spread from the lowest to the greatest value in the distribution is indicated by the range.
The calculation of this variability index is the simplest.
Simply subtract the lowest value from the highest value in the data set to determine the range.

How do you find the range?

The difference between the lowest and highest values in a list or set is known as the range.
Put all the numbers in order before determining the range. The lowest number should then be subtracted from the highest.
The range of the list is provided in the response.

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1 year ago

Evaluate the interval (Calculus 2)

Darya [45]

Answer:

$2 \tan (6x)+2 \sec (6x)+\text{C}$

Step-by-step explanation:

<u>Fundamental Theorem of Calculus</u>

$\displaystyle \int \text{f}(x)\:\text{d}x=\text{F}(x)+\text{C} \iff \text{f}(x)=\dfrac{\text{d}}{\text{d}x}(\text{F}(x))$

If differentiating takes you from one function to another, then integrating the second function will take you back to the first with a constant of integration.

Given indefinite integral:

$\displaystyle \int \dfrac{12}{1-\sin (6x)}\:\:\text{d}x$

$\boxed{\begin{minipage}{5 cm}\underline{Terms multiplied by constants}\\\\$\displaystyle \int a\:\text{f}(x)\:\text{d}x=a \int \text{f}(x) \:\text{d}x$\end{minipage}}$

If the terms are multiplied by constants, take them outside the integral:

$\implies 12\displaystyle \int \dfrac{1}{1-\sin (6x)}\:\:\text{d}x$

Multiply by the conjugate of 1 - sin(6x) :

$\implies 12\displaystyle \int \dfrac{1}{1-\sin (6x)} \cdot \dfrac{1+\sin(6x)}{1+\sin(6x)}\:\:\text{d}x$

$\implies 12\displaystyle \int \dfrac{1+\sin(6x)}{1-\sin^2(6x)} \:\:\text{d}x$

$\textsf{Use the identity} \quad \sin^2 x+ \cos^2 x=1:$

$\implies \sin^2 (6x) + \cos^2 (6x)=1$

$\implies \cos^2 (6x)=1- \sin^2 (6x)$

$\implies 12\displaystyle \int \dfrac{1+\sin(6x)}{\cos^2(6x)} \:\:\text{d}x$

Expand:

$\implies 12\displaystyle \int \dfrac{1}{\cos^2(6x)}+\dfrac{\sin(6x)}{\cos^2(6x)} \:\:\text{d}x$

$\textsf{Use the identities }\:\: \sec \theta=\dfrac{1}{\cos \theta} \textsf{ and } \tan\theta=\dfrac{\sin \theta}{\cos \theta}:$

$\implies 12\displaystyle \int \sec^2(6x)+\dfrac{\tan(6x)}{\cos(6x)} \:\:\text{d}x$

$\implies 12\displaystyle \int \sec^2(6x)+\tan(6x)\sec(6x) \:\:\text{d}x$

$\boxed{\begin{minipage}{5 cm}\underline{Integrating $\sec^2 kx$}\\\\$\displaystyle \int \sec^2 kx\:\text{d}x=\dfrac{1}{k} \tan kx\:\:(+\text{C})$\end{minipage}}$

$\boxed{\begin{minipage}{6 cm}\underline{Integrating $ \sec kx \tan kx$}\\\\$\displaystyle \int \sec kx \tan kx\:\text{d}x= \dfrac{1}{k}\sec kx\:\:(+\text{C})$\end{minipage}}$

$\implies 12 \left[\dfrac{1}{6} \tan (6x)+\dfrac{1}{6} \sec (6x) \right]+\text{C}$

Simplify:

$\implies \dfrac{12}{6} \tan (6x)+\dfrac{12}{6} \sec (6x)+\text{C}$

$\implies 2 \tan (6x)+2 \sec (6x)+\text{C}$

Learn more about indefinite integration here:

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3 0

2 years ago