Answer:
Three hundred and fifty-four thousand, two hundred and ten =354210
Hundred thousand║Ten thousand║Thousand║Hundred║Tens║Ones
3 5 4 2 1 0
Step-by-step explanation:
Answer:
576 yards
Step-by-step explanation:
Since the integers are consecutive, you could call them 'x', x+1,and x+2.
Or you could all them x-1, 'x', and x+1.
Or you could call them x-2, x-1, and 'x'.
It's up to you. Each of those sets describes 3 consecutive numbers.
Most people will call them 'x', x+1, and x+2, so let's do that.
Now just go through the question and make the quantities it talks about:
"twice the first" . . . . 2(x)
"three times the second" . . . . 3(x+1)
"four times the third" . . . . . 4(x+2)
and finally, it says that the sum of those things is 20.
This should not be a major problem.
2x + 3(x+1) + 4(x+2) = 20
You can probably handle it from here, but I'll continue anyway.
The next thing to do is take out all the parentheses:
3(x+1) = 3x + 3
4(x+2) = 4x + 8
Now add up the whole thing again, but without parentheses:
2x + 3x + 3 + 4x + 8 = 20
Gather all the x's together, and all the naked numbers together:
(2x + 3x + 4x) + (3 + 8) = 20
Addum up: 9x + 11 = 20
Subtract 11 from each side: 9x = 9
Divide each side by 9 : x = 1
The first integer = x . . . . . . . <em>1</em>
The second integer = x+1 . .<em> . 2</em>
The third integer = x+2 . . . . .<em> 3 </em>
<u>Check:</u>
"twice the first" . . . . . . . . . . 2(1) = 2
"three times the second". . . 3(2) = 6
"four times the third" . . . . . . 4(3) = 12
Addum up: 2 + 6 + 12 = 20 just like the question said. yay!
Answer:
y = x² + 7x + 4
Step-by-step explanation:
<u>First option</u>
System of equations:
y = x² + x – 4
y = x – 5
Replacing:
x - 5 = x² + x – 4
0 = x² + x – 4 - x + 5
0 = x² + 1
Discriminant:
b²-4*a*c
0²-4*1*1
-4 < 0 then the equation has no real roots
<u>Second option</u>
System of equations:
y = x² + 2x – 1
y = x – 5
Replacing:
x - 5 = x² + 2x – 1
0 = x² + 2x - 1 - x + 5
0 = x² + x + 4
Discriminant:
b²-4*a*c
1²-4*1*4
-15 < 0 then the equation has no real roots
<u>Third option</u>
System of equations:
y = x² + 6x + 9
y = x – 5
Replacing:
x - 5 = x² + 6x + 9
0 = x² + 6x + 9 - x + 5
0 = x² + 5x + 4
Discriminant:
b²-4*a*c
5²-4*1*4
9 > 0 then the equation has two different real roots
<u>Fourth option</u>
System of equations:
y = x² + 7x + 4
y = x – 5
Replacing:
x - 5 = x² + 7x + 4
0 = x² + 7x + 4 - x + 5
0 = x² + 6x + 9
Discriminant:
b²-4*a*c
6²-4*1*9
0 then the equation has one real root
