the complete question in the attached figure
Volume of the sand V=(1/3)*h*π*r²
we have that
h=30 mm
r=10 mm
v=(1/3)*30*π*10²=1000≈3140 mm³
the rate is 50 mm³/seg
50 mm³--------------------------------------> 1 seg
3140 mm³-------------------------------- X
x=3140/50=62.8 seg
the answer is 62.8 seg
33)31
34) 31
35) 21
Check the last one to make sure it’s right . The first 2 I’m 100% sure they are correct
Answer:
Slope-intercept
y = 3/4(x) - 7
Point slope
y -5= 3/4(x - 16)
Step-by-step explanation:
In slope-intercept
We have the general slope intercept as;
y = mx + b
where m is the slope and b is the y-intercept
in this case, m = 3/4 and b = -7
So we have;
y = 3/4(x) - 7
In point-slope
we have the general form as;
y-y1 = m(x-x1)
So what we have is as follows;
y -5= 3/4(x - 16)
Where we have (x1,y1) = (16,5)
the answer is b because x is greater than or equal to 0
X^2 = 3x + 5
x^2 - 3x - 5 = 0
a(alpha) and b(beta are roots
(x - a)(x - b) = 0
x^2 - (a+b)x + ab = 0
compare coefficients
a+b = 3
ab = -5
solving
(1/a^2 + 1/b^2)
= (a^2 + b^2)/(ab)^2
= (a^2 + 2ab + b^2 - 2ab)/(ab)^2
= [(a+b)^2 - 2ab]/(ab)^2
= [(3)^2 - 2(-5)]/(-5)^2 =...
prove
a^4 = 57a + 70
let x=a,
a^2 = 3a + 5
(a^2)^2 = (3a + 5)^2
a^4 = 9a^2 + 30a + 25
a^4 = 9(3a + 5) + 30a + 25
a^4 = 57a + 70
