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3 years ago

How many times does 0.4 fl oz go into 60 ml

1 answer:

6 0

1 oz= 29.5735296 ml

Convert ounces to milliliters by multiplying milliliters in 1 ounce by 0.4 ounces.

=0.4 oz * 29.5735296 ml/oz
=11.829 ml

Divide 60 ml by the portion size (11.829 ml)

=60 ÷ 11.829
= 5.07 times

ANSWER: 5.07 times

Hope this helps! :)

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the cost of fencing a rectangular field at rupees 30 per metre is rupees 2400 if the length of the its field is 24m than its bre

prohojiy [21]

Answer:

breadth = 8, length = 32

Step-by-step explanation:

perimeter = 2 x (length x breadth)

= 2 x ((24 + b) + b))

= 2 x (24 + 2b)

= 48 + 4 b

48 + 4b meters costs rupees 2400

each meter costs rupees 30

so,

30 x (48 + 4b) = 2400

1440 + 120b = 2400

120b = 960

b = 8

the breadth is 8m, the length is 8 + 24 = 32m

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3 years ago

What value of c makes the polynomial below a perfect square? x^2+10+c

Inessa05 [86]

Answer: Hi there

To find the last term in order to obtain the perfect square, take the middle term, divide it by 2, and take its square.

10 ÷ 2 = 5

5^2 = 25

Thus, the equation would be x^2 + 10x + 25

The answer is 25

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3 years ago

Espen predicted that he would sell 32 baseball caps, but he actually sold 29 baseball caps. Which expression would find the perc

tiny-mole [99]

Answer:C is the answer

Step-by-step explanation:

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3 years ago

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Jackie found the area of a square to be 25/16 square feet.Which shows the side length of the square?

svlad2 [7]

The side length is the square root of the area, so is ...

... √(25/16 ft²) = (√25)/(√16) ft = 5/4 ft

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3 years ago

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Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false:

kondaur [170]

$\text{Proof by induction:}$
$\text{Test that the statement holds or n = 1}$

$LHS = (3 - 2)^{2} = 1$
$RHS = \frac{6 - 4}{2} = \frac{2}{2} = 1 = LHS$
$\text{Thus, the statement holds for the base case.}$

$\text{Assume the statement holds for some arbitrary term, n= k}$
$1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2} = \frac{k(6k^{2} - 3k - 1)}{2}$

$\text{Prove it is true for n = k + 1}$
$RTP: 1^{2} + 4^{2} + 7^{2} + ... + [3(k + 1) - 2]^{2} = \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2} = \frac{(k + 1)[6k^{2} + 9k + 2]}{2}$

$LHS = \underbrace{1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2}}_{\frac{k(6k^{2} - 3k - 1)}{2}} + [3(k + 1) - 2]^{2}$
$= \frac{k(6k^{2} - 3k - 1)}{2} + [3(k + 1) - 2]^{2}$
$= \frac{k(6k^{2} - 3k - 1) + 2[3(k + 1) - 2]^{2}}{2}$
$= \frac{k(6k^{2} - 3k - 1) + 2(3k + 1)^{2}}{2}$
$= \frac{k(6k^{2} - 3k - 1) + 18k^{2} + 12k + 2}{2}$
$= \frac{k(6k^{2} - 3k - 1 + 18k + 12) + 2}{2}$
$= \frac{k(6k^{2} + 15k + 11) + 2}{}$
$= \frac{(k + 1)[6k^{2} + 9k + 2]}{2}$
$= \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2}$
$= RHS$

Since it is true for n = 1, n = k, and n = k + 1, by the principles of mathematical induction, it is true for all positive values of n.

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3 years ago