What is the slope of the tangent of the function y = x square + X + 4 at point (- 1,4)?

What is the slope of the tangent of the function y = x square + X + 4 at point (- 1,4)?

y=x^2+x+4
y'=2x+1
Substitute - 1 to get
Y '= - 1, so the slope is - 1
The tangent equation is y-4 = - (x + 1)

Given that the slope of the tangent of the function FX = ax + CX (a, C ∈ R) at point (3,6) is 8, find the analytic expression of the function FX
emergency

f(x)=ax³+cx
f'(x)=3ax²+c
f'(3)=8=27a+c
f(3)=6=27a+3c
The solution is a = 1 / 3, C = - 1
f(x)=x³/3-x

The tangent slope of F (x) = ln (2-x) + ax at point (0, f (0)) is 1 / 2

F '(x) = - 1 / (2-x) + A, k = f' (0) = - 1 / 2 + a = 1 / 2, a = 1
So f (x) = ln (2-x) + X (x)

1-2 / 3x = 3x-2 / 5
process

1-2 / 3x = 3x-2 / 5
The result of the transfer is as follows:
5 / 1 + 2 = 3x-3x / 2
3x / 2 = 7 / 2
X = 7 / 2 ／ 3 / 2
X = 7 / 3

(1) X times 1 / 7 + 5 = 37 (2) 8x-12.2 = 11.8 (3) X-1 / 3 = 3x + 1 / 5

(1) X times 1 / 7 + 5 = 37
1/7x=37-5
1/7x=32
x=32x7
x=224
（2）8x-12.2=11.8
8x=11.8+12.2
8x=24
x=3
(3) X-3 / 1 x = 5 / 3 x + 1
2/3x-3/5x=1
1/15x=1
x=15
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Solution equation: - 2 / 1 x + 3 X-1 = 3 / 1 x + 2

-1 / 2 x + 3 X-1 = 1 / 3 x + 2
Multiply both sides by 6 to get
-3x+18x-6=2x+12
13x=18
x=18/13

Solution equation: 3 (x-21) = 2x + 3x

3x-63=2x+3x
2x=-63
x=-63/2=-31.5

(3x-2) × 7 = 2x (35-21) solution equation

(3x-2)×7=2x(35-21)
21x-14=28x
28x-21x=-14
7x=-14
x=-2

Simple calculation of 0.11 * 28.2-0.72 * 1.1-0.11

0.11*28.2—0.72*1.1—0.11
=1·1÷10×2·82×10－0.72*1.1—1·1×0·1
=1·1×2·82－0.72*1.1—1·1×0·1
=1·1×（2·82－0.72－0·1）
=1·1×2
=2·2

Can 64.51-1 / 11 + 1 / 11 * 3 + 1 / 2 be simplified?

64.51-1/11+1/11*3+1/2
=64.51+2/11+1/2
=65.01+2/11
=65+1/100+2/11
=65+(11+200)/1100
=65 and 211 / 1100