Point slope equation:
y-y1=m(x-x1)
m=slope
So we simply plug in our given information:
x1=-8
y1=2
m=1/2
y-2=1/2(x-(-8)
2 minus signs next to each other make a positive
Final answer:
y-2=1/2(x+8)
6. One variable only so pretty straightforward.
length-x+4
width-x
x+x+4=80
2x=76
x=38
x+4=42
answer: length 42cm and width 38cm
7. Another money problem!
n-# of nickels
q-# of quarters
n=3+q
0.05n+0.25q=2.85
Substitution works like a charm!
0.05(3+q)+0.25q=2.85
0.15+0.05q+0.25q=2.85
0.3q=2.7
q=9
n=3+q
n=3+9
n=12
answer: 9 nickels and 12 quarters
8. One variable situation again.
Ann's money-2b+9
Betty's money-b
b+2b+9=60
3b=51
b=17
2b+9=43
answer: Ann has $43 and Betty has $17.
9. # of red m&m's-x+1
# of blue m&m's-x
x+1+x=13
x=6
x+1=7
answer: 6 blue and 7 red m&m's
10. a-number of adult tickets
s-number of student tickets
a+s=785 ----> a=785-s
5a+2s=3280
5(785-s)+2s=3280
-3s=-645
s=215
a+s=785
a+215=785
a=570
answer: 215 children tickets and 570 adult tickets
Answer:
x=8+1/2
Step-by-step explanation:
Hope this helps, good luck!
<span>A new kind of temporary pavilion support for a square roof uses just two poles set at the diagonal corners of a square. The allowable distance between the poles is 18 feet. Find the area of the roof. The sides will have lengths of 18/sqrt(2). Since they are perpendicular, the area of the square will be (18^2)/2 = 81*4/2 = 162 square feet.
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The distance between the point (3, 5) and a line with the equation 5x – 3y + 10 = 0 will be 10 / √(34) units.
<h3>What is the distance between a point and a line?</h3>
Let (x₁, y₁) be the point and ax + by + c = 0 be the line.
Then the distance between a point and a line will be
The point (3, 5) and a line with the equation 5x – 3y + 10 = 0.
Then the distance between them will be
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