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Mademuasel [1]
3 years ago
6

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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3x + 2 3 y - 15 &lt; 0 The point (x, 9) is a solution for the inequality shown. What is a possible value for x?
love history [14]

Answer:

x is any number less than -64; i.e. -66

Step-by-step explanation:

Inequality

3x + 23y - 15 < 0

where point (x,9) satisfies this

Substitute 9 for y

3x + 23(9) - 15 < 0

3x + 207 - 15 < 0

3x + 192 < 0       Take away 192 from both sides

3x < -192             Divide both sides by 3 to isolate x

x < -64


Check:

3(-64) + 23(9) -15 < 0

-192 + 207 - 15 < 0

-192 + 192 < 0

Therefore x must be less that -64 to satisfy the point (x, 9) in this inequality.

4 0
3 years ago
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Which two whole numbers does 7 1/2 lie between
goldfiish [28.3K]

Hi there! :)

<u>Answer:</u>

7 1/2 lies between the two whole numbers <u>7</u> and <u>8</u>.


Step-by-step explanation:

"<u>Whole numbers</u>" are numbers without fractional or decimal parts and no negatives.

The first whole number <u>BEFORE</u> 7 1/2 is 7.

The first whole number <u>AFTER</u> 7 1/2 is 8.


There you go! I really hope this helped, if there's anything just let me know! :)

4 0
3 years ago
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ExtremeBDS [4]

The square root of 7 would like between 2 and 3. This is because 2 x 2 is 4's square root and 3 x 3 is nine's square root. Seven is between four and nine.

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Murljashka [212]
I believe the answer is vertical stretch
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) dy 2x<br> ------ = ---------------<br> dx yx2 + y
larisa86 [58]

Step-by-step explanation:

\dfrac{dy}{dx} = \dfrac{2x}{y(x^2 + 1)}

Rearranging the terms, we get

ydy = \dfrac{2xdx}{x^2 + 1}

We then integrate the expression above to get

\displaystyle \int ydy = \int \dfrac{2xdx}{x^2 + 1}

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or

y = \sqrt{2\ln |x^2 + 1|} + k

where I is the constant of integration.

6 0
3 years ago
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