Answer:
80, 96, 8, 10
Step-by-step explanation:
The first set gives us the ratio, 32:2 which is basically 16
5*16= 80
6*16= 96
Notice how it is on the top now, so divide by 16
128/16= 8
160/16=10
Answer:
Present age
Son: 15
Father: 45
Step-by-step explanation:
Remark
Thank you for the translation. Without it, the problem would be impossible -- at least for me.
Givens
Let the present age of the father = x
Let the present age of the son = y
Solution
x + y = 60
How many years will pass? You could say it's z.
x + z = y When z years pass, the son will be his father's present age.
x + z + y + z = 120 when z is added to both their current ages, the result is 120 Collect like terms
x + y + 2z = 120
<u>x + y = 60 </u> Subtract The very first equation
2z = 60 Divide by 2
z = 60/2 30 years have passed.
z = 30
x + z = y
x + 30 = y Substitute x + 30 for the present y value (the father).
x + x + 30 = 60
2x = 30
x = 15
x + y = 60
15 + y = 60
y = 60 - 15
y = 45
So the son's age right now is 15
The father's age right now is 45
Let's use the point-slope formula...
<em>y-y1=m(x-x1)
</em>
<em>m is the slope.
</em>
<em>y--9=7(x--2)
</em>
<em>Subtracting a negative number is the same as adding a positive number...
</em>
<em>y+9=7(x+2)
</em>
<em>y+9=7x+14
</em>
<em>Let's subtract 9 from both sides...
</em>
<em>-9+y+9=7x+14-9
</em>
<em>y=7x+5
</em>
<em>The formula is now in the format of...
</em>
<em>y=mx+b
</em>
<em>This is known as slope-intercept.
</em>
<em>m is the slope.
</em>
<em>b is the y-intercept, the value of y when x=0.
</em>
<em>Standard formula is...
</em>
<em>Ax+By=C
</em>
<em>Neither A nor B equal zero.
</em>
<em>A is greater than zero.
</em>
<em>y=7x+5
</em>
<em>Let's move 7x to the left side of the equation. It becomes negative.
</em>
<em>-7x+y=5
</em>
<em>Let's multiply both sides by -1 to render A greater than zero.
</em>
<em>-1(-7x+y)=(5)(-1)
</em>
<u>7x-y=-5
</u>
<u>This is the equation in standard form.</u>
Answer:
The Range of possible measures for the third side is; 2 < C < 16
Step-by-step explanation:
In a triangle with sides A , B and C , the sum of the length of any two sides of a triangle must be greater than the third side:
Given the two sides of a triangle are 7 and 9.
Let A = 7 and B = 9 then;
B-A< C < A+B
Then; the range of C can be:
9-7 < C < 7+9
Simplify:
2 < C < 16
Therefore, the range of possible measures for the third side is; 2 < C < 16
