2^(1+2n)
2^(1+0) = 2
2^(1+2) = 2^3 = 8
2^(1+4) = 2^5 = 32
One side of the rectangle is x=2, the other side is 2x-5
Add up all the four sides: (x+2) +(x+2)+(2x-5)+(2x-5)=54
6x-6=54
x=10
#3: suppose the first integer is x, then the second one is x+2
x(x+2)=255
x^2 +2x -255 =0
factor the quadratic equation: (x+17)(x-15)=0
x=-17, which is impossible, or x=15
so the two positive integers are 15 and 17
Answer:
The answer is C.
Step-by-step explanation:
Given formula h(t)=−16t2+v0t+h0 , where v0 is the initial velocity and h0 is the initial height.
In this case, the initial postion is a platform 30ft above ground so h0=+30
The initial velocity is 38 ft/s straight up into the air so v0=+38
h(t)=-16t2+38t+30
When the object hits the ground, h=0.
h=-16t2+38t+30=0
Simplifying 8t2-19t-15=0
(8t+5)(t-3)=0
t=-5/8 or 3
As time cannot be -ve, t=3s. The answer is C.
Answer:
b = 159 m
a = 103 m
angle B is 57 degrees
Step-by-step explanation:
Let’s start with getting the measure of angle B
Mathematically, since we have a right triangle , we have to subtract the given angle from 90 to get the other acute angle
So we have the measure of B as;
90-33 = 57 degrees
To get the value of a, we use the appropriate trigonometric ratio
a faces the angle 33, that makes it an opposite to that angle
190 faces the right angle and that makes it the hypotenuse
Mathematically , the relationship between the two is that trigonometrically, they are related by the sine which is the ratio of the opposite to the hypotenuse
so, we have it that;
sine 33 = a/190
a = 190 * sine 33
a = 103 m
to get b, it is adjacent to the given angle
So with the hypotenuse, the ratio between it and the hypotenuse is the cosine
so;
cos 33 = b/190
b = 190 * cos 33
b = 159 m
Nathan= 1/10 of the job per day
Matt= 1/15 of the job per day
3/30 + 2/30. =5/30
1/(5/30) = 30/5. Or 6 days for both working together.
