A lorry weighs 4.65 tonnes. It carries a load of 748 kg. What is the total weight of the lorry and its load?

Mathematics

2 answers:

Lesechka [4]3 years ago

7 0

5472.61813 kilograms

evablogger [386]3 years ago

3 0

(4.65 tonnes)+(748 kg) =5.39800 tonnes

You might be interested in

How did I get a 2 here (simplifying ratios)

PilotLPTM [1.2K]

Answer:

$\frac{2x {}^{2} + 14x + 24 }{4 {x}^{2} + 4x - 48 } = \frac{x + 4}{2(x - 2)}$

Step-by-step explanation:

Please see attached picture for full solution.

4 0

3 years ago

What is the range of the function for the given domain? y = 2x − 5 D: {-2, 0, 2, 4}

rosijanka [135]

Answer:

Range of Function : { - 9, - 5, - 1, 4 }

Step-by-step explanation:

We know that y = 2x - 5 provided the domain ( x - values ) { - 2, 0, 2, 4 }. Let us substitute each element in this set of domain as x in the equation "y = 2x - 5" as to solve for the y - values, otherwise known as the range of the function.

{ - 2, 0, 2, 4 }

y = 2( - 2 ) - 5 = - 9,

y = 2( 0 ) - 5 = - 5,

y = 2( 2 ) - 5 = - 1,

y = 2( 4 ) - 5 = 4

We have the set of y - values as { - 9, - 5, - 1, 4 }. This is the range of our function.

7 0

3 years ago

Read 2 more answers

How do you find the derivative of 1/logx?

valentina_108 [34]

The derivative of 1/logx is With the chain rule.

1log(x)=log(x)−1 is ,= -1xlog(x)2 .

The by-product of logₐ x (log x with base a) is 1/(x ln a). Here, the thrilling issue is that we have "ln" withinside the by-product of "log x". Note that "ln" is referred to as the logarithm (or) it's miles a logarithm with base "e".

The by-product of 1/log x is -1/x(log x)^2. Note that 1/logx is the reciprocal of log.

$With the chain rule. \\ 1log(x)=log(x)−1 \\ So you follow the chain rule to the electricity after which to the log: \\ ddxlog(x)−1=−1⋅log(x) \\ \\−2⋅ddxlog(x) \\ =−1log(x)21x \\ =−1xlog(x)2 .$