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2 years ago

Tom works for a company his normal rate of pay 15 per hour

Mathematics

2 answers:

Valentin [98]2 years ago

8 0

The total pay Tom earn for working 11 hours is £185

Given that Tom is paid £15 if he works for less than 7 hours and then paid 1 1/3 times his normal salary if he works more than 7 hours.

Given that he works for 11 hours, hence:

Pay for the first 7 hours = £15 × 7 hours = £105

Pay for the overtime = 1 1/3 × £15 × (11 - 7) = 4/3 × £15 × 4 = £80

Total pay = £80 + £105 = £185

Hence the total pay Tom earn for working 11 hours is £185

Find out more at: brainly.com/question/21938284

Brrunno [24]2 years ago

3 0

The amount earned by Tom is the product of the rate per hour and the number of hours worked which is $90.

The amount earned per hour = $15

The number of hours worked = 6 hours

The amount earned at the normal earning rate can be calculated thus :

Normal rate per hour × number of hours

Amount earned = $15 × 6 = $90

Therefore, the amount earned by Tom after working for 6 hours at the normal rate is $90.

Learn more : brainly.com/question/18796573

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If 3x^2 + y^2 = 7 then evaluate d^2y/dx^2 when x = 1 and y = 2. Round your answer to 2 decimal places. Use the hyphen symbol, -,

S_A_V [24]

Taking $y=y(x)$ and differentiating both sides with respect to $x$ yields

$\dfrac{\mathrm d}{\mathrm dx}\bigg[3x^2+y^2\bigg]=\dfrac{\mathrm d}{\mathrm dx}\bigg[7\bigg]\implies 6x+2y\dfrac{\mathrm dy}{\mathrm dx}=0$

Solving for the first derivative, we have

$\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{3x}y$

Differentiating again gives

$\dfrac{\mathrm d}{\mathrm dx}\bigg[6x+2y\dfrac{\mathrm dy}{\mathrm dx}\bigg]=\dfrac{\mathrm d}{\mathrm dx}\bigg[0\bigg]\implies 6+2\left(\dfrac{\mathrm dy}{\mathrm dx}\right)^2+2y\dfrac{\mathrm d^2y}{\mathrm dx^2}=0$

Solving for the second derivative, we have

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=-\dfrac{3+\left(\frac{\mathrm dy}{\mathrm dx}\right)^2}y=-\dfrac{3+\frac{9x^2}{y^2}}y=-\dfrac{3y^2+9x^2}{y^3}$

Now, when $x=1$ and $y=2$ , we have

$\dfrac{\mathrm d^2y}{\mathrm dx^2}\bigg|_{x=1,y=2}=-\dfrac{3\cdot2^2+9\cdot1^2}{2^3}=\dfrac{21}8\approx2.63$

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First, distribute the 2 in the left side of the equation.
Resulting in: 2x-6 = (x-1) + 7

Second, remove the parentheses form the right side of the equation (there is nothing to distribute there).
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Now simplify the right side of the equation by subtracting the 1 from the 7.
Resulting in: x + 6

Our goal is to isolate the x to the left side, and the numerals to the right side.
With that in mind, add the 6 (from the right side) to both sides. This cancels out the 6 on the right side.
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Lastly, in order to fully isolate the variable (x), we divide both sides by 2.
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