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3 years ago

Is 8,456mL less than 9L

Mathematics

2 answers:

nikitadnepr [17]3 years ago

6 0

8,456mL is smaller than 9L because 9L equals 9000mL

dalvyx [7]3 years ago

4 0

Hello,

Here is your answer:

The proper answer to this question is "yes"! One liter is 1,000 mL!

Your answer is yes!

If you need anymore help feel free to ask me!

Hope this help!

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Correlation of a scatter plot question in image

Ierofanga [76]

Answer:

For this question there appears to be absolutely no association. The points are all over the place and there is not consistent factors at play here.

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3 years ago

How many solutions does this equation have?

dedylja [7]

Answer:

3n = 3 - 3n

<=>

6n = 3

<=>

n = 2

=> This equation has one solution.

6 0

3 years ago

Read 2 more answers

Please help me fast and show all your work please thx

Kitty [74]

If the two diagonals are $d$ and $D$ , the are of the rhombus is

$A=\dfrac{dD}{2}$

So, in your case, the area is

$A=\dfrac{15\cdot 35}{2}=262.5$

The other options are wrong because:

option A doens't divide by 2, thus getting twice the area of the rhombus
option B summed the diagonals instead of multiplying them
option D took the difference of the diagonals.

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3 years ago

Pharmacist used 826 milliliters of a liquid for an experiment. How many liters of the liquid did the scientist use for this expe

liberstina [14]

Answer: 0.826 litres
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7 0

2 years ago

Find the minimum value of the function f(x)=x^2+5x-6

mash [69]

<h3><u>Explanation</u></h3>

Method 1 (Formula)

$\begin{cases}h = - \frac{b}{2a} \\ k = \frac{4ac - {b}^{2} }{2a} \end{cases}$

The vertex of Parabola is the maximum/minimum point depending on the value of a.

Find Vertex

<u>h-value</u>

$h = - \frac{5}{2(1)} \\ h = - \frac{5}{2}$

<u>k-value</u>

$k = \frac{4(1)( - 6) - {(5)}^{2} }{4(1)} \\ k = \frac{ - 24 - 25}{4} \\ k = \frac{ - 49}{4} \\ k = - \frac{49}{4}$

The minimum value is the value of k. Therefore the minimum value is - 49/4 at x = -5/2.

Method 2 (Derivative)

This is Calculus method. We simply differentiate the function then substitute y' = 0.

Differentiate Function

$f'(x) = 2 {x}^{2 - 1} + {5x}^{1 - 1} - 0 \\ f'(x) = 2x + 5$

Substitute f'(x) = 0

$0 = 2x + 5 \\ - 5 = 2x \\ - \frac{5}{2} = x$

Substitute x = -5/2 in the original equation.

$f(x) = {( - \frac{5}{2} )}^{2} + 5( - \frac{5}{2} ) - 6 \\ f(x) = \frac{25}{4} - \frac{25}{2} - 6 \\ f(x) = \frac{25}{4} - \frac{50}{4} - \frac{24}{4} \\ f(x) = \frac{25}{4} - \frac{74}{4} \\ f(x) = - \frac{49}{4}$

<h3><u>Answer</u><u /></h3>

<u> $\sf{the \: \: minimum \: \: value \: \: is \: \: - \frac{49}{4} \: \: at \: \: x = - \frac{5}{2} }$ </u>

4 0

3 years ago