Answer:
the digit 3 (or 30,000,000,000) is in the ten billions place.
What deos the triangle look like and what are the answer chioces
Answer:
f(x)=-18x^2
Step-by-step explanation:
Given:
1+Integral(f(t)/t^6, t=a..x)=6x^-3
Let's get rid of integral by differentiating both sides.
Using fundamental of calculus and power rule(integration):
0+f(x)/x^6=-18x^-4
Additive Identity property applied:
f(x)/x^6=-18x^-4
Multiply both sides by x^6:
f(x)=-18x^-4×x^6
Power rule (exponents) applied"
f(x)=-18x^2
Check:
1+Integral(-18t^2/t^6, t=a..x)=6x^-3
1+Integral(-18t^-4, t=a..x)=6x^-3
1+(-18t^-3/-3, t=a..x)=6x^-3
1+(6t^-3, t=a..x)=6x^-3
That looks great since those powers are the same on both side after integration.
Plug in limits:
1+(6x^-3-6a^-3)=6x^-3
We need 1-6a^-3=0 so that the equation holds true for all x.
Subtract 1 on both sides:
-6a^-3=-1
Divide both sides by-6:
a^-3=1/6
Raise both sides to -1/3 power:
a=(1/6)^(-1/3)
Negative exponent just refers to reciprocal of our base:
a=6^(1/3)
Your answer will be D. Hope I helped.
Let us say that:
a = ones
b = fives
c = twenties
So that the total money is:
1 * a + 5 * b + 20 * c = 229
=> a + 5b + 20c = 229 -->
eqtn 1
We are also given that:
c = a – 5 -->
eqtn 2
a + b + c = 30 -->
eqtn 3
Rewriting eqtn 3 in terms of b:
b = 30 – a – c
Plugging in eqtn 2 into this:
b = 30 – a – (a – 5)
b = 35 – 2a -->
eqtn 4
Plugging in eqtn 2 and 4 into eqtn 1:
a + 5(35 – 2a) + 20(a – 5) = 229
a + 175 – 10a + 20a – 100 = 229
11a = 154
a = 14
So,
b = 35 – 2a = 7
c = a – 5 = 9
Therefore there are 14 ones, 7 fives, and 9 twenties.
