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Anton [14]
4 years ago
14

what is the shape of a cross section that is perpendicular to the base of a cube A. rectangle B. triangle C.square D. circle

Mathematics
1 answer:
777dan777 [17]4 years ago
8 0

Answer:

2qwertfghtrew3eret

Step-by-step explanation:

3ewrew

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A math tutor earns $40 an hour tutoring teenagers. Sometimes he will meet with two students at once and get paid $80 per hour to
kati45 [8]

Answer:

It should be linear if you let the variable t represent the time tutoring for ONE student,   so   y = 40*t   where t = number of hours for ONE student

so  if the tutor taught  2 students for 2 hours together, then t = 2*2 = 4 hours

Step-by-step explanation:

y = 40t

t = tutoring hours for ONE unique student...

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35/7=5
4/5
A: 4/5
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Gambar grafik fungsi kuadrat dari y=x²-x-6
sveticcg [70]
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4 0
3 years ago
A line tangent to the curve f(x)=1/(2^2x) at the point (a, f(a)) has a slope of -1. What is the x-intercept of this tangent?
kirza4 [7]

Answer:

x-intercept = 0.956

Step-by-step explanation:

You have the function f(x) given by:

f(x)=\frac{1}{2^{2x}}   (1)

Furthermore you have that at the point (a,f(a)) the tangent line to that point has a slope of -1.

You first derivative the function f(x):

\frac{df}{dx}=\frac{d}{dx}[\frac{1}{2^{2x}}]  (2)

To solve this derivative you use the following derivative formula:

\frac{d}{dx}b^u=b^ulnb\frac{du}{dx}

For the derivative in (2) you have that b=2 and u=2x. You use the last expression in (2) and you obtain:

\frac{d}{dx}[2^{-2x}]=2^{-2x}(ln2)(-2)

You equal the last result to the value of the slope of the tangent line, because the derivative of a function is also its slope.

-2(ln2)2^{-2x}=-1

Next, from the last equation you can calculate the value of "a", by doing x=a. Furhtermore, by applying properties of logarithms you obtain:

-2(ln2)2^{-2a}=-1 \\\\2^{2a}=2(ln2)=1.386\\\\log_22^{2a}=log_2(1.386)\\\\2a=\frac{log(1.386)}{log(2)}\\\\a=0.235

With this value you calculate f(a):

f(a)=\frac{1}{2^{2(0.235)}}=0.721

Next, you use the general equation of line:

y-y_o=m(x-x_o)

for xo = a = 0.235 and yo = f(a) = 0.721:

y-0.721=(-1)(x-0.235)\\\\y=-x+0.956

The last is the equation of the tangent line at the point (a,f(a)).

Finally, to find the x-intercept you equal the function y to zero and calculate x:

0=-x+0.956\\\\x=0.956

hence, the x-intercept of the tangent line is 0.956

5 0
3 years ago
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