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:
54
Step-by-step explanation:
I'm assuming the equation looks like this: 15x - 6
15*4 - 6
60 - 6
54
OR if it's just x-6 then the answer is -2
Hope this is what you were asking, have a nice day! :)
Step-by-step explanation:
by using similarity and enlargement
10/8 = 7+y/7
70=56+8y
8y=14
y= 7/4
10/8 = 5+x/5
50=40+8x
x = 8/10 = 4/5
1/sin^2x-1/tan^2x=
1/sin^2x-1/ (sin^2x/cos^2x)<<sin tan= sin/cos>>
= 1/sin^2x- cos^2x / sin^2x
= (1- cos^2x) / sin^2x <<combining into a single fraction>>
sin^2 x / sin^2x <<since 1- cos^2 x sin^2 x
=1
this simplifies to 1.
