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3 years ago

In a 180-kilogram sample of ore, there was 3.2% metal. How many kilograms of metal were in the sample?

Mathematics

1 answer:

shtirl [24]3 years ago

7 0

5.76 KG as 180 x 0.032 (3.2 percent) = 5.76

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What if it is that cool pirate money that is shaped weird? I hear it is 11/12 foot long and ¾ wide. What is the square foot of t

cestrela7 [59]

If you are asking how to find the square feet of the "weirdly shaped pirate money" you'd need to multiply the length and width. So:

11/12 * 3/4 = 33/48

So the answer is 33/48 square feet

If you are not asking for this answer word your question better please.

8 0

3 years ago

What is the difference of (6d+2)-(3d -3)

lapo4ka [179]

Answer:

3d+5

Step-by-step explanation:

Say x is the difference

Solve for x.

6d+2−(3d−3)=x

Step 1: Flip the equation.

x=6d+2−(3d−3)

Step 2: Simplify

x=(6d-3d)+(2+3)

x=3d+5

7 0

3 years ago

If R is the midpoint of OS , RS = 2x - 4, ST = 4x – 1, and RT = 8x – 43, find QS.

RoseWind [281]

Answer:

Step-by-step explanation:

5 0

3 years ago

Insert 3 rational numbers between

avanturin [10]

Answer:

$\frac{ \sqrt{20} }{20} \\ \sqrt{ \frac{1}{4} \times \frac{ \sqrt{20} }{20} } \\ \sqrt{ \frac{1}{5} \times \frac{ \sqrt{20} }{20} }$

Step-by-step explanation:

If you have two numbers x and y, their geometric mean is defined as

$\sqrt{xy}$

Let that number be m, and let x be the smallest of the two numbers.

You know for sure that

$x \leqslant m \leqslant y$

Since

$m \geqslant \sqrt{ {x}^{2} }$

And

$m \leqslant \sqrt{ {y}^{2} }$

With equality holding if and only if x=y.

So, we can use the geometric mean to generate numbers which are for sure between two desired values. For example:

$\frac{1}{4} < \sqrt{ \frac{1}{4} \times \frac{1}{5} } < \frac{1}{5}$

The middle term is of course irrational, since it's equal to

$\frac{1}{ \sqrt{20} } = \frac{ \sqrt{20} }{20}$

Now we can do the same thing with our newfound value which we know to be between 1/4 and 1/5:

$\frac{1}{4} < \sqrt{ \frac{1}{4} \times \frac{ \sqrt{20} }{20} } < \frac{1}{5}$

And

$\frac{1}{4} < \sqrt{ \frac{1}{5} \times \frac{ \sqrt{20} }{20} } < \frac{1}{5}$

8 0

3 years ago

For how many two-digit numbers is the tens digit larger than the ones digit

Mamont248 [21]

<span>ALL IN ALL THERE ARE 45 DIFFERENT TENS DIGIT LARGER THAN THE ONES</span>

Download xlsx

5 0

4 years ago