What is the slope of the tangent at point (0,2) of the function y = ex plus 1

What is the slope of the tangent at point (0,2) of the function y = ex plus 1

y=e^x
Derivation
The slope of Y '= e ^ x at (0,2) is e ^ 0 = 1
therefore
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Let the function f (x) = 4x3 + ax + 2, and the slope of the tangent of the curve y = f (x) at point P (0,2) be - 12. Find the value of a [4x3 is the power of 4 times x] Please talk about the process

f(x)'=12x2+a
Because at point P (0,2), f (o) '= - 12
So bring x = 0 into the first line
A = - 12
The slope is a derivative of the function

Find: (1 + x) power 5 = 10. What is x equal to? How much is x equal to (1 + x) power 5 = 10? I hope to write the calculation process and answer

(1+x)^5=10
1+x=5√10=10^0.2
1+x≈1.5848931924611134852021013733915
x≈0.5848931924611134852021013733915

Given that the x-th power of 2 is equal to the y-th power of 5 is equal to 10, what is one of x plus one of Y

2^x=5^y=10
xlg2=ylg5=1
x=1/lg2 y=1/lg5
1/x=lg2 1/y=lg5
1/x+1/y=lg2+lg5=lg10=1

The 10th power of x equals 1.5. What is x? RT

It is 1.041379743992410586841910102311

The x power of 2 is equal to 3, and the Y power of 2 is equal to 5. Divide the x power of 4 by the Y power of 8

The x power of 2 is equal to 3, the Y power of 2 is equal to 5,
Divide the x power of 4 by the Y power of 8
=(2^x)^2÷(2^y)^3
=3^2÷5^3
=9/125;
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