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3 years ago

8 1/2 x 2 3/4 x 2/3= 12

Mathematics

1 answer:

Aloiza [94]3 years ago

7 0

Answer:

155.25= 12

Step-by-step explanation:

17/2*2*1/2 = 12

17*1/2 = 12

8 1/2 17/1* 1/2

8*2+1 17*1/1*2

2 17*1/1*2

16+1 17/1*2

2 17/2

17 8.5=12

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2 years ago

What is the inverse of the given relation? y=3x+12

daser333 [38]

We are asked in the problem to determine the inverse of the relation y = 3x + 12. first step is to express the equation in terms of y, that is y -12 = 3x, then exchange the places of x and y, that is x - 12 = 3y. This is the final answer

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2 years ago

Read 2 more answers

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erik [133]

Answers:

When we evaluate a logarithm, we are finding the exponent, or power x, that the base b, needs to be raised so that it equals the argument m. The power is also known as the exponent.

$5^2 = 25 \to \log_5(25) = 2$

The value of b must be positive and not equal to 1

The value of m must be positive

If 0 < m < 1, then x < 0

A logarithmic equation is an equation with a variable that includes one or more logarithms.

===============================================

Explanation:

Logarithms, or log for short, basically undo what exponents do.

When going from $5^2 = 25$ to $\log_5(25) = 2$ , we have isolated the exponent.

More generally, we have $b^x = m$ turn into $\log_b(m) = x$

When using the change of base formula, notice how

$\log_b(m) = \frac{\log(m)}{\log(b)}$

If b = 1, then log(b) = log(1) = 0, meaning we have a division by zero error. So this is why $b \ne 1$

We need b > 0 as well because the domain of y = log(x) is the set of positive real numbers. So this is why m > 0 also.

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2 years ago

Suppose you want to create a set of weights so that any object with an integer weight from 1 to 40 pounds can be balanced on a t

Archy [21]

Answer:

You need 4 weights, and the weights are 1; 3; 9 and 27.

Step-by-step explanation:

Since the scale has two plates, we can place weights on either side and also the object so it can be balanced.

This is a key part of the problem, it's not only on the other side of the scale, but on both sides.

Let's do the math now.

If i get two weights, 1 and 3. I can form this combinations.

Object of 1lb = 1

Object of 2lb + 1 weight = 3 weight.

Object of 3lb = 3 weight

Object of 4lb = 1 weight + 3 weight.

So what if i want to add the next weight and that weight to add me the maximum amount of objects. The weight would have to have a difference with the last object plus one. So if i grab 9. 9 minus 4 is 5. And that is a difference with the last object plus 1.

With a weight of 9, now i can add all the integers up to 13lb.

And the next step? Lets add one more. Keeping the last rule, the weight would have to have a difference with the last object plus one. So if i grab 27, 27 minus 13 is 14. And that is a difference witht the last object plus 1.

The sum of all the weights adds up to 40 pounds. And i can balance any integer in the middle.

The formula we are using is p – n = n + 1

Where p is the new weight. and n is the last object we weighted. And the sum of the weights goes up to the last object we can place on the scale, and in this case is 40.

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3 years ago

-6=-8+ <img src="https://tex.z-dn.net/?f=%20%5Cfrac%7Bp%7D%7B11%7D%20" id="TexFormula1" title=" \frac{p}{11} " alt=" \frac{p}

vekshin1

P=22 rewrite the question as -8+p/11=-6 then move all terms not containing p to the right side p/11=2 multiply both sides of the given equation by 11, the reciprocal, 11*p/11=11*2

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2 years ago