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3 years ago

How many millimetres are there in fifteen centimetres

Mathematics

2 answers:

Anika [276]3 years ago

8 0

150 (in one centimeter there is 10 millimeters multiply 10 times 15 you get 150)

Anastaziya [24]3 years ago

3 0

<span>When solving a question like this, with two quantities and their different units (centimetres and millimetres in this case) you can begin by converting the values to the same units. this makes the calculations easier.

15 centimeters to millimeters = 15 x 100 ( because 1 cm has 100mm)

There are 1500 mm in 15 centimetres.</span>

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inysia [295]

Answer:

D. The graph of function $g$ is the graph of function $f$ shifted 4 units to the left.

Step-by-step explanation:

the function $g$ is composite function between $f$ and $x + 4$ , then you can re-write $g$ as:

$g(x) = 10^{x+4}$

The transformation happened in the input, then you have an horizontal shifted and how it's adding 4 units then you go "faster" and move to left the graph.

In the image you can see what's mean sum a constant in the $x$ of a function.

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2 years ago

Verify by substitution whetherthe given functions are solutions of the given DE. Primes denote derivatives with respect to x.y!!

julia-pushkina [17]

Complete Question

The complete question is shown on the first uploaded

Answer:

$y_1$ is not a solution of the differential equation

$y_2$ is not a solution of the differential equation

$y_3$ is not a solution of the differential equation

Step-by-step explanation:

The differential equation given is $y'' + y' = cos2x$

Let consider the first equation to substitute

$y_1 = cosx +sinx$

$y_1' = -sinx +cosx$

$y_1'' = -cosx -sinx$

$y_1'' - y_1' = -cosx -sinx -sinx +cosx$

$y_1'' + y_1' = -2sinx$

$-2sinx \ne cos2x$

This means that $y_1$ is not a solution of the differential equation

Let consider the second equation to substitute

$y_2 = cos2x$

$y_2' = -2sin2x$

$y_2'' = -4cos2x$

$y_2'' + y_2' = -4cos2x-2sin2x$

$-4cos2x-2sin2x \ne cos2x$

This means that $y_2$ is not a solution of the differential equation

Let consider the third equation to substitute

$y_3 = sin 2x$

$y_3' = 2cos 2x$

$y_3'' = -4sin2x$

$y_3'' + y_3' = -4sin2x - 2cos2x$

$-4sin2x - 2cos2x \ne cos2x$

This means that $y_3$ is not a solution of the differential equation

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3 years ago

Plz help me with the questions for math

Sonja [21]

Answer:

2. 6/9

3.2/4

4. 8/10

Step-by-step explanation:

basically the idea is to find something in the same fact family so like the example on 1, follow that and try to answer the next problams on your own its really easy

( sorry if im wrong or not i tried to help)

7 0

3 years ago

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FromTheMoon [43]

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