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Sergio [31]
3 years ago
6

Quadratic Word Problems

Mathematics
1 answer:
Anit [1.1K]3 years ago
5 0
4. Let the numbers be x - 2, x, and x + 2, where x is an odd number.
2(x² - 4) - 4x = (x + 2)² + 21
2x² - 8 - 4x = x² + 4x + 4 + 21
2x² - 4x - 8 = x² + 4x + 25
x² - 8x - 33 = 0
(x - 11)(x + 3) = 0

x = 11, or -3
When x = 11, x - 2 = 9, x + 2 = 13

When x = -3, x - 2 = -5, x + 2 = -1

Explanation: You are on the right track. However, rather than having three unknown variables, try to reduce your working out to one unknown variable. Since you know they are consecutive odd numbers, you can simply let x be the middle term and the other two be + and - 2, provided x is an odd number.

That will reduce your variable issues, and helps as the first and third provide a difference of two squares, and this works out very nicely.

Q5: is essentially the same process. Let your variables be something in the form of one unknown variable, and you should be okay from there. Let me know if you're stuck.
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36.75 divided by 508.25<br> then round to the nearest tenth
gavmur [86]

Answer:

0.1

Step-by-step explanation:

Before rounding: 0.07230693556

After rounding: 0.1

Hope this helps!!!!

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3 years ago
Classify What type of number is the square root of 2 ​
Amanda [17]

Answer:

The square root of 2, or the (1/2)th power of 2, written in mathematics as √2 or 2​1⁄2, is the positive algebraic number that, when multiplied by itself, equals the number 2. Technically, it is called the principal square root of 2, to distinguish it from the negative number with the same property

Step-by-step explanation:

8 0
3 years ago
) f) 1 + cot²a = cosec²a​
notsponge [240]

Answer:

It is an identity, proved below.

Step-by-step explanation:

I assume you want to prove the identity. There are several ways to prove the identity but here I will prove using one of method.

First, we have to know what cot and cosec are. They both are the reciprocal of sin (cosec) and tan (cot).

\displaystyle \large{\cot x=\frac{1}{\tan x}}\\\displaystyle \large{\csc x=\frac{1}{\sin x}}

csc is mostly written which is cosec, first we have to write in 1/tan and 1/sin form.

\displaystyle \large{1+(\frac{1}{\tan x})^2=(\frac{1}{\sin x})^2}\\\displaystyle \large{1+\frac{1}{\tan^2x}=\frac{1}{\sin^2x}}

Another identity is:

\displaystyle \large{\tan x=\frac{\sin x}{\cos x}}

Therefore:

\displaystyle \large{1+\frac{1}{(\frac{\sin x}{\cos x})^2}=\frac{1}{\sin^2x}}\\\displaystyle \large{1+\frac{1}{\frac{\sin^2x}{\cos^2x}}=\frac{1}{\sin^2x}}\\\displaystyle \large{1+\frac{\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}}

Now this is easier to prove because of same denominator, next step is to multiply 1 by sin^2x with denominator and numerator.

\displaystyle \large{\frac{\sin^2x}{\sin^2x}+\frac{\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}}\\\displaystyle \large{\frac{\sin^2x+\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}

Another identity:

\displaystyle \large{\sin^2x+\cos^2x=1}

Therefore:

\displaystyle \large{\frac{\sin^2x+\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}\longrightarrow \boxed{ \frac{1}{\sin^2x}={\frac{1}{\sin^2x}}}

Hence proved, this is proof by using identity helping to find the specific identity.

6 0
3 years ago
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tangare [24]

Answer:

1/54

Step-by-step explanation:

1/9 x 1/6

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3 years ago
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