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2 years ago

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What weight of dry substance is in 150g of a 3% substance solution? What weight of an 8% solution can we have with the same weig

ht of dry substance?
Answer:

g of dry substance and
g of an 8% solution.

1 answer:

Pavlova-9 [17]2 years ago

5 0

The answers are: 4.5 g and 56.25 g respectively.

Since the first type of measurement in this question is weight or mass, I'll suppose that the percentage concentration is % mass/mass. For that type of concentration measurement, just multiply the percentage by the total mass to get the mass of the wanted material.
So 150 g * 3% = 150 g * 0.03 = 4.5g

For the 8% solution with the same amount of dry substance, use the ratio of percentages, multiplied by the mass of the first solution to get the wanted amount of new solution:
3/8 * 150 g = 56.35 g

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Sadie simplified the expression StartRoot 54 a Superscript 7 b cubed EndRoot, where a greater-than-or-equal-to 0, as shown colon

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Answer:

Sadie's error is " she made error in step 2 $=\sqrt{3^2\times 6\times a^2\times a^5\times b^2\times b}$ where $a\geqslant 0$ "

Because she made error in splitting the powers to simplify the square root

<h3>Therefore the correct answer for Sadie's expression is $3ab\sqrt{6ab}$ where $a\geqslant 0$ </h3>

Step-by-step explanation:

Given that " Sadie simplified the expression StartRoot 54 a Superscript 7 b cubed EndRoot, where a greater-than-or-equal-to 0, "

It can be written as $\sqrt{54a^7b^3}$ where $a\geqslant 0$

The given expression is $\sqrt{54a^7b^3}$ where $a\geqslant 0$

To find Sadie's error and explain the correct answer :

Sadie's steps are

$\sqrt{54a^7b^3}$ where $a\geqslant 0$

$=\sqrt{3^2\times 6\times a^2\times a^5\times b^2\times b}$

$=3ab\sqrt{6a^5b}$

<h3> $\sqrt{54a^7b^3}=3ab\sqrt{6a^5b}$ where $a\geqslant 0$ </h3><h3><u>Now corrected steps are</u></h3>

$\sqrt{54a^7b^3}$ where $a≥0$

$=\sqrt{(9\times 6)(a^{6+1})(b^{2+1})$

$=\sqrt{(3^2\times 6)(a^6.a^1)(b^2.b^1)$ (by using the identity $a^{m+n}=a^m.a^n$

$=\sqrt{3^2\times 6\times ((a^3)^2.a)(b^2.b)$ (by using the identity $a^{mn}=(a^m)^n$ )

$=3ab\sqrt{6ab}$

Therefore $\sqrt{54a^7b^3}=3ab\sqrt{6ab}$ where $a\geqslant 0$

<h3>The correct answer is $3ab\sqrt{6ab}$ where $a\geqslant 0$ </h3>

Sadie's error is " she made error in step 2 $=\sqrt{3^2\times 6\times a^2\times a^5\times b^2\times b}$ " where $a\geqslant 0$

Because she made error in splitting the powers to simplify the square root

<h3>Therefore the correct answer for Sadie's expression is $3ab\sqrt{6ab}$ where $a\geqslant 0$ </h3>

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2 years ago

Read 2 more answers

Let P and Q be polynomials with positive coefficients. Consider the limit below. lim x→[infinity] P(x) Q(x) (a) Find the limit i

jenyasd209 [6]

Answer:

If the limit that you want to find is $\lim_{x\to \infty}\dfrac{P(x)}{Q(x)}$ then you can use the following proof.

Step-by-step explanation:

Let $P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0}$ and $Q(x)=b_{m}x^{m}+b_{m-1}x^{n-1}+\cdots+b_{1}x+b_{0}$ be the given polinomials. Then

$\dfrac{P(x)}{Q(x)}=\dfrac{x^{n}(a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n})}{x^{m}(b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m})}=x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}$

Observe that

$\lim_{x\to \infty}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\dfrac{a_{n}}{b_{m}}$

and

$\lim_{x\to \infty} x^{n-m}=\begin{cases}0& \text{if}\,\, nm\end{cases}$

Then

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Step-by-step explanation:

The data we have here is:

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