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pantera1 [17]
3 years ago
6

If you put 25ml into a glass how much water is needed

Mathematics
1 answer:
dlinn [17]3 years ago
6 0

Answer: it is depends on what unit of measurement your look looking for

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18 2by5-4 1by10 find the difference
luda_lava [24]

Given:

The expression is:

18\dfrac{2}{5}-4\dfrac{1}{10}

To find:

The difference.

Solution:

We have,

18\dfrac{2}{5}-4\dfrac{1}{10}

It can be rewritten as:

=\dfrac{90+2}{5}-\dfrac{40+1}{10}

=\dfrac{92}{5}-\dfrac{41}{10}

Taking LCM, we get

=\dfrac{184-41}{10}

=\dfrac{143}{10}

=\dfrac{140+3}{10}

=14\dfrac{3}{10}

Therefore, the value of the difference is 14\dfrac{3}{10}.

3 0
3 years ago
Which math function describes the following situation?
MissTica
This is the concept of inequality;
We are told that the distance a car travels depends on how full gas tank is, this means that the car will travel until the point when gasoline is finished. Thus,
This mean that if the if the distance is given by the function f(x), x will be the gasoline since it is the dependent variable, liters this in other words can be written as (if the tank holds 89 ):

f(x)=distance(gasoline)<89

thus the correct answer will be:

B] Distance (amount of gasoline) <  89

4 0
3 years ago
Find the simplified product<br> ^3V9x^4 x ^3V3x^8
Kay [80]

\sqrt[3]{9x^4}\cdot \sqrt[3]{3x^8}\implies \sqrt[3]{(9x^4)(3x^8)}\implies \sqrt[3]{27x^{4+8}}\implies \sqrt[3]{27x^{12}} \\\\\\ \sqrt[3]{3^3(x^4)^3}\implies 3x^4

8 0
2 years ago
Nancy was keeping track of how many miles she ran each month. In April, she ran 25 miles. In May, she ran 34 miles, and in June,
Elanso [62]

Answer:

She ran 101 miles within the 3 months. an estimate would be around 100

4 0
3 years ago
find the slope of the curve y=x^2-2x-5 at the point P(2,5) by finding the limit of secant slopes through point P
Fynjy0 [20]

The point (2, 5) is not on the curve; probably you meant to say (2, -5)?

Consider an arbitrary point Q on the curve to the right of P, (t,y(t))=(t,t^2-2t-5), where t>2. The slope of the secant line through P and Q is given by the difference quotient,

\dfrac{(t^2-2t-5)-(-5)}{t-2}=\dfrac{t^2-2t}{t-2}=\dfrac{t(t-2)}{t-2}=t

where we are allowed to simplify because t\neq2.

Then the equation of the secant line is

y-(-5)=t(x-2)\implies y=t(x-2)-5

Taking the limit as t\to2, we have

\displaystyle\lim_{t\to2}t(x-2)-5=2(x-2)-5=2x-9

so the slope of the line tangent to the curve at P as slope 2.

- - -

We can verify this with differentiation. Taking the derivative, we get

\dfrac{\mathrm dy}{\mathrm dx}=2x-2

and at x=2, we get a slope of 2(2)-2=2, as expected.

4 0
3 years ago
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