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Lelu [443]
3 years ago
8

For every pound a company spends on advertising, it spends £0.43 on its website. Express the amount spent on advertising to its

website as a ratio in its simplest form.
Mathematics
1 answer:
Wittaler [7]3 years ago
6 0

Answer:

£1 : £0.43

Step-by-step explanation:

So to make the ratio, you need the amount of money spent on advertising and the money spent on its website

Website money = £0.43

Advertising money = £1

£1 : £0.43

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line m passes through point (-2,-1) and is perpendicular to the graph of y=-2/3x+6. Line n is parallel to line m and passes thro
puteri [66]

Answer:

Hello...... here is a solution :

1 -  

Hello:

the   equation of "m" is : y = ax+b

the slope is a : a×(- 2/3) = -1......( perpendicular to a line : y=-2/3x+6 when the slope is -2/3 )

a = 3/2          

the line " n" that passes through (4, - 3) and  parallel to line "m" :  

when the slope is 3/2 ( same slope )

the equation in slope-intercept form of the line" n" is :

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7 0
3 years ago
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The equation x squared + y squared = r squared
WITCHER [35]
(5-0)^2+(-2+0)^2=r^2\\
25+4=r^2\\
29=r^2\\\\
\boxed{x^2+y^2=\sqrt{29}}
5 0
3 years ago
Use the equation a = IaIâ
german

Answer:

a) \:\:=\sqrt{14}\cdot \frac{\:\:}{\sqrt{14} }

b)\:\:=\sqrt{29} \cdot \frac{\:\:}{\sqrt{29} }

c) \:\:=7\cdot \frac{\:\:}{7}

Step-by-step explanation:

a) Let <u>a</u>=<2,1,-3>

The magnitude of <u>a</u> is |a|=\sqrt{2^2+1^2+(-3)^2}

|a|=\sqrt{4+1+9}=\sqrt{14}

The unit vector in the direction of a is

\hat{a}=\frac{\:\:}{\sqrt{14} }

Using the relation a=|a|\hat{a}, we have

\:\:=\sqrt{14}\cdot \frac{\:\:}{\sqrt{14} }

b) Let a=2i - 3j + 4k

|a|=\sqrt{2^2+(-3)^2+4^2}

|a|=\sqrt{4+9+16}=\sqrt{29}

\hat{a}=\frac{\:\:}{\sqrt{29} }

Using the relation a=|a|\hat{a}, we have

\:\:=\sqrt{29} \cdot \frac{\:\:}{\sqrt{29} }

c) Let us first find the sum of <1, 2, -3> and <2, 4, 1> to get:

<1+2, 2+4, -3+1>=<3, 6, -2>

Let a=<3, 6, -2>

The magnitude is

|a|=\sqrt{3^2+6^2+(-2)^2}

|a|=\sqrt{9+36+4}=\sqrt{49}=7

The unit vector in the direction of <u>a</u> is

\hat{a}=\frac{\:\:}{7}

Using the relation a=|a|\hat{a}, we have

\:\:=7\cdot \frac{\:\:}{7}

5 0
3 years ago
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