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4 years ago

8

Give one example of when you would. use the friendly numbers strategy to subtract .Explain why ?

1 answer:

lord [1]4 years ago

5 0

Say you were solving 27 + 19. If you were planning on using the friendly numbers strategy, you'd look at the first <span>addend, 27, and determine that</span> the closest "friendly number" here is thirty, which you'd get by adding 3. From there, you can jump to the next friendly number, forty, by adding another 10. So far you'd've added 13, and you'd need another 6 for that too add up to 19, your second addend. After adding 3, 10, and 6 to 27, you'd find your sum, or "final answer"- 46.

I hope this was what you were looking for!

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Number 6 and 7 in fraction form and decimal

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4 years ago

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A positive real number is 2 less than another. When 4 times the larger is added to the square of the smaller, the result is 49.

Kobotan [32]

Answer:

The numbers are

$-2+3\sqrt{5}$ and $3\sqrt{5}$

Step-by-step explanation:

Let

x -----> the smaller positive real number

y -----> the larger positive real number

we know that

A positive real number is 2 less than another

so

$x=y-2$

$y=x+2$ ----> equation A

When 4 times the larger is added to the square of the smaller, the result is 49

so

$4y+x^2=49$ ----> equation B

substitute equation A in equation B

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solve for x

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The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$x^2+4x-41=0$

so

$a=1\\b=4\\c=-41$

substitute in the formula

$x=\frac{-4\pm\sqrt{4^{2}-4(1)(-41)}} {2(1)}$

$x=\frac{-4\pm\sqrt{180}} {2}$

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$x=-2\pm3\sqrt{5}$

so

The positive real number is

$x=-2+3\sqrt{5}$

Find the value of y

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$y=-2+3\sqrt{5}+2$

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4 years ago

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2 years ago

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zhuklara [117]

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