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Lerok [7]
3 years ago
10

Find the tangent line to the graph of f(x)=x at point (0,0)

Mathematics
1 answer:
ratelena [41]3 years ago
4 0

Given:

The function is

f(x)=x

To find:

The tangent line to the graph of f(x)=x at point (0,0).

Solution:

We have,

f(x)=x

It is a linear function because the highest power of the variable is 1.

We know that the tangent line to a linear function at any point is the line itself.

Derivative of given function is

f'(x)=1

So, slope of the tangent line is 1. It is given table the tangent line passes through the point (0,0). The equation of tangent line is

y-0=1(x-0)

y=x

Therefore, the tangent line to the graph of f(x)=x at point (0,0) is y=x.

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5 0
2 years ago
Write each number in standard form 6.7x10 to the 1 power
e-lub [12.9K]
<h2>Answer: 67</h2>

Step-by-step explanation: 6.7 x 10^1 = 67

8 0
3 years ago
Which graph shows the solution to the system of linear inequalities?<br> y&lt; 2x-5<br> y&gt;-3x + 1
trapecia [35]

Answer:

See explanation

Step-by-step explanation:

Plot the solution sets to both inequalities.

1. For the inequality y First, plot the dotted line y=2x+5 (dotted because sign is without notion "or equal to"), then choose correct part by substitution coordinates of the origin.

2\cdot 0-5=-5

so the origin does not belong to the needed part. Shade the part, which does not include origin.

2. For the inequality y>-3x+1. First, plot the dotted line y=-3x+1 (dotted because sign is without notion "or equal to"), then choose correct part by substitution coordinates of the origin.

-3\cdot 0+1=1>0

so the origin does not belong to the needed part. Shade the part, which does not include origin.

3. Find the common region of these two shaded parts - this is the solution to the system of two inequalities.

5 0
3 years ago
in a AP the first term is 8,nth term is 33 and sum to first n terms is 123.Find n and common difference​
allsm [11]

I believe there is no such AP...

Recursively, this sequence is supposed to be given by

\begin{cases}a_1=8\\a_k=a_{k-1}+d&\text{for }k>1\end{cases}

so that

a_k=a_{k-1}+d=a_{k-2}+2d=\cdots=a_1+(k-1)d

a_n=a_1+(n-1)d

33=8+(n-1)d

21=(n-1)d

n has to be an integer, which means there are 4 possible cases.

Case 1: n-1=1 and d=21. But

\displaystyle\sum_{k=1}^2(8+21(k-1))=37\neq123

Case 2: n-1=21 and d=1. But

\displaystyle\sum_{k=1}^{22}(8+1(k-1))=407\neq123

Case 3: n-1=3 and d=7. But

\displaystyle\sum_{k=1}^4(8+7(k-1))=74\neq123

Case 4: n-1=7 and d=3. But

\displaystyle\sum_{k=1}^8(8+3(k-1))=148\neq123

8 0
3 years ago
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