What a delightful little problem ! (Partly because I could see
right away how to do it, and had the answer in a few minutes,
after a lot of impressive-looking algebra on my scratch-paper.)
Three consecutive integers are . . . x, x+1, and x+2
The smallest two are . . . x and x+1
Their product is . . . . . x(x+1)
5 times the largest one is . . . 5(x+2)
5 less than that is . . . . . . 5(x+2)-5
Now, the conditions of the problem say that <u>x (x + 1) = 5 (x+2) - 5</u>
THAT's the equation we have to solve, to find 'x' .
Eliminate parentheses: x² + x = 5x + 10 - 5
Combine like terms: x² + x = 5x + 5
Subtract 5x from each side: x² - 4x = 5
Subtract 5 from each side: <u>x² - 4x - 5 = 0</u>
You could solve that by factoring it, or use the quadratic equation.
Factored, it says that (x + 1) (x - 5) = 0
From which <em>x = -1</em>
and <em>x = +5</em>
We only want the positive results, so our three consecutive integers are
5, 6, and 7 .
To answer the question, the smallest one is <em><u>5 </u></em>.
<u>Check</u>:
5 x 6 ? = ? (7 x 5) - 5
30 ? = ? (35) - 5
30 = 30
yay !
Answer:
2
Step-by-step explanation:
The range of class A is 5-0 = 5
The range of class B is 3-0 = 3
the difference between 3 and 5 is 2.
The tangent secant theorem says
PR × PS = PQ²
4 × (4+5) = PQ²
36 = PQ²
PQ=6
Answer: 6
Answer:
(x-1)(x+7)
Step-by-step explanation:
x = 1
x = -7
These are the roots found using the quadratic formula.
Answer:
--- 1 over 5 squared
Step-by-step explanation:
When multiplying terms with a common base, you just add the exponents:
That's true even when you don't have any exponents.
A negative exponent isn't fully simplified, so there's another rule to use:
That is '1 over x to the y' if it's too small to read.
