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fredd [130]
3 years ago
6

The average weight of 8 marbles is 3.5 g. What is the total weight of the marbles?

Mathematics
1 answer:
professor190 [17]3 years ago
8 0

Answer:

Take 8÷3.5. So the answer is aprominently 2.29.

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Solve |3x + 3) = 15.
prisoha [69]
It’s B, because you do 3-15 and that equals 12 and then you do 3 divided by 12 and that’s 4.
8 0
2 years ago
Read 2 more answers
a metallurgist has an alloy with 10% titanium and an alloy with 20% titanium. he needs 100 grams of an alloy with 17% titanium.
oee [108]

Let us say that:

x = mass of 10% titanium alloy required

y = mass of 20% titanium alloy required

To solve this problem, we set-up two mass balances or equations.

Overall mass balance:    x + y = 100

                                                x = 100 – y                           ---> 1

Titanium mass balance: 0.10 x + 0.20 y = 0.17 (100)

                                                0.10 x + 0.20 y = 17           ---> 2

Now we combine equations 1 and 2:

0.10 (100 – y) + 0.20 y = 17

10 – 0.10 y + 0.20 y = 17

0.10 y = 7

y = 70 g

 

Calculating for x using equation 1:

x = 100 – y

x = 100 – 70

x = 30 g


Answer:

x = 30 g of 10% titanium alloy required

y = 70 g <span>of 20% titanium alloy required</span>

6 0
3 years ago
You are making punch for a baby shower. Your make 6 quarts of punch. How many pints of punch did you make?
3241004551 [841]
You made 12 pints of punch
8 0
3 years ago
The cost of 5 pairs of shoes is $108.36. What is the unit rate for cost per<br> pair?
zysi [14]

Answer: 21.67 and full Answer is 21.6732

Step-by-step explanation: $108.36 divide by 5!

8 0
3 years ago
Suppose the roots of the equation 2x^2−5x−6=0 are α and β.Find the quadratic equation with roots 1/α and 1/β.​
igor_vitrenko [27]

The quadratic equation with roots 1/α and 1/β is 6 x^{2}+5 x-2=0

<u>Solution:</u>

Given, roots of equation 2 x^{2}-5 x-6=0 \text { are } \alpha, \beta

We have to find equation whose roots are 1/α, and 1/β.

We know that,

\text { sum of roots }=\frac{-x \text { cosficient }}{x^{2} \text { coefficient }} \rightarrow \alpha+\beta=\frac{-(-5)}{2} \rightarrow \alpha+\beta=\frac{5}{2}

\text { And, product of roots }=\frac{c o n s t a n t}{x^{2} c o e f f i c i e n t} \rightarrow \alpha \beta=\frac{-6}{2} \rightarrow \alpha \beta=-3

Now, we know that, general form of an equation is

x^{2}-(\text { sum of roots }) x+\text { product of roots }=0

Then, equation whose roots 1/α and 1/β is

\begin{array}{l}{x^{2}-\left(\frac{1}{\alpha}+\frac{1}{\beta}\right) x+\frac{1}{\alpha} \times \frac{1}{\beta}=0} \\\\ {\rightarrow x^{2}-\left(\frac{\alpha+\beta}{\alpha \beta}\right) x+\frac{1}{\alpha \beta}=0}\end{array}

from above given equation,

\begin{array}{l}{\rightarrow x^{2}-\left(\frac{\frac{5}{2}}{-3}\right) x+\frac{1}{-3}=0} \\\\ {\text { multiplying equation with }-3} \\\\ {\rightarrow-3 x^{2}-\frac{5}{2} x+1=0} \\\\ {\text { multiplying equation with } 2} \\\\ {\rightarrow-6 x^{2}-5 x+2=0} \\\\ {\rightarrow-6 x^{2}-5 x+2=0} \\\\ {\text { multiplying equation with }-1} \\\\ {\rightarrow 6 x^{2}+5 x-2=0} \\\\ {\text { Hence, the required line equation is } 6 x^{2}+5 x-2=0}\end{array}

4 0
3 years ago
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