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3 years ago

how much of a 10% syrup solution should be added to 2 liters of a 30% syrup solution to get 20% syrup solution?

1 answer:

4 0

<span>10x + 30(2) = 20(x + 2)
10x + 60 = 20x + 40
10x = 20
answer
x = 2 liters
So you would need 2 liters of the 10% solution to make the 30% solution 20%.</span>

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Which expression is equivalent to \frac{10q^5w^7}{2w^3}.\ \frac{4\left(q6\right)^2}{w^{-5}} for all values of q and w where the

mestny [16]

Answer:

$\frac{10q^5w^7}{2w^3}.\ \frac{4\left(q^6\right)^2}{w^{-5}} =20q^{17}w^9$

Step-by-step explanation:

Given

$\frac{10q^5w^7}{2w^3}.\ \frac{4\left(q^6\right)^2}{w^{-5}}$

Required

Determine the equivalent expression

$\frac{10q^5w^7}{2w^3}.\ \frac{4\left(q^6\right)^2}{w^{-5}}$

Simplify the first fraction

$\frac{5q^5w^7}{w^3}.\ \frac{4\left(q^6\right)^2}{w^{-5}}$

Apply law of indices on the first fraction;

$5q^5w^{7-3}.\ \frac{4\left(q^6\right)^2}{w^{-5}}$

$5q^5w^4.\ \frac{4\left(q^6\right)^2}{w^{-5}}$

$\ \frac{5q^5w^4*4\left(q^6\right)^2}{w^{-5}}$

Apply law of indices:

$5q^5w^4*4\left(q^6\right)^2 * w^{5}$

Evaluate the bracket

$5q^5w^4*4 * q^{12} * w^{5}$

Collect Like Terms

$5*4q^5* q^{12}*w^4 * w^{5}$

$20q^5* q^{12}*w^4 * w^{5}$

$20q^{5+12}*w^{4+5}$

$20q^{17}w^9$

Hence:

$\frac{10q^5w^7}{2w^3}.\ \frac{4\left(q^6\right)^2}{w^{-5}} =20q^{17}w^9$

7 0

3 years ago

Pls answer this question cuz i don't know what is the answer on this one

Sveta_85 [38]

Answer:

12.D

13.C

14.A

15.B

16.A

17.B

18.D

19.A

20.A

and hey I don't know the answer to 11 th question as I don't know much about miles

PS: Have a nice day :)

6 0

3 years ago

Read 2 more answers

Please help if u cannot skip

posledela

The statement that is true about the function is D. it is discontinuous and non-differentiable at x = 3.

<h3>How to determine which statement is true?</h3>

To determine which statement is true, we need to know the conditions for continuity and differentiablity of a function.

<h3>Conditions for continuity and differentiablity of a function.</h3>

For a function f(x) to be continuous at a point x = a, then both the left hand limit of f(x) and the right hand limit of f(x) as x → a must be equal. That is $\lim_{x \to a^{-} } f(x) = \lim_{x \to a^{+} } f(x)$ . So, $\lim_{x \to a^{} } f(x)$ must exist since $\lim_{x \to a^{-} } f(x) = \lim_{x \to a^{+} } f(x) = \lim_{x \to a^{} } f(x)$

Also, for a function to be differentiable at a point x = a, it must also exist at x = a

So, since f(x) = {x² - 1 if -1 ≤ x ≤ 3 and x²/3 if 3 < x ≤ 8}

From the equality on the first condition,we see that f(x) is exists at x = 3 but is not continuous since f(x) changes to another function when x > 3. So,left hand limit of f(x) and the right hand limit of f(x) as x → 3 are not equal.

That is $\lim_{x \to 3^{-} } f(x) \neq \lim_{x \to 3^{+} } f(x)$ . Thus, the function is discontinuous at x = 3.