6x+5y=-15
5y=-15-6x
y=(-15-6x)/5
X=9
A) 3x-7=34
3x= 34+7
3x= 41
x= 41/3
B) x-9=1
x=1+9
x=10
C) dude what is this
D) 4x+5=41
4x=41-5
4x=36
x=36/4
x=9 YAYY
Hope it helps!
#MissionExam001
Taking the derivative of 7 times secant of x^3:
We take out 7 as a constant focus on secant (x^3)
To take the derivative, we use the chain rule, taking the derivative of the inside, bringing it out, and then the derivative of the original function. For example:
The derivative of x^3 is 3x^2, and the derivative of secant is tan(x) and sec(x).
Knowing this: secant (x^3) becomes tan(x^3) * sec(x^3) * 3x^2. We transform tan(x^3) into sin(x^3)/cos(x^3) since tan(x) = sin(x)/cos(x). Then secant(x^3) becomes 1/cos(x^3) since the secant is the reciprocal of the cosine.
We then multiply everything together to simplify:
sin(x^3) * 3x^2/ cos(x^3) * cos(x^3) becomes
3x^2 * sin(x^3)/(cos(x^3))^2
and multiplying the constant 7 from the beginning:
7 * 3x^2 = 21x^2, so...
our derivative is 21x^2 * sin(x^3)/(cos(x^3))^2
A circle with a radius of 4cm sits inside a circle with a radius of 11cm
What is the area of the shaded region?
Round your final answer to the nearest hundredth.
Answer: 329.86 cm
Answer:
4 seconds.
Step-by-step explanation:
The function f(x)=-10(x)(x-4) ........ (1), represents the approximate height of a projectile launch on the ground into the air as a function of time in seconds x.
Now, we are asked that for how long from the launch does the projectile stays in the air.
Therefore, we have to solve the equation (1) making f(x) as zero.
Hence, 10x(x - 4) = 0
⇒ x = 0 or x = 4
(As x can not be zero since at x = 0 sec, the projectile was at the ground.}
Hence, x = 4 seconds.
Therefore, the projectile was in the air for 4 seconds. (Answer)
