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sergij07 [2.7K]
3 years ago
11

What is the domain of this scatter plot?

Mathematics
1 answer:
natka813 [3]3 years ago
8 0

Answer:

this answer should be green

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Could someone please solve this and explain how to do it?
galben [10]

Answer:

no clue buddy sry

Step-by-step explanation:

4 0
3 years ago
Find the slope of the tangent line to the curve f(x)=e^(x) at (0.4,1.49)
almond37 [142]

Answer:

1.49

Step-by-step explanation:

In order to find the slope of the tangent line to a given equation, and in a given point, we need to:

1. Find the first derivative of the given function.

2. Evaluate the first derivative function in the given point.

1. Let's find the first derivative of the given function:

The original function is f(x)=e^{x}

But remeber that the derivative of  e^{x} is  e^{x}

so, f'(x)=e^{x}

2. Let's evaluate the first derivative function in the given point

The given point is (0.4,1.49) so:

f'(x)=e^{x}

f'(0.4)=e^{0.4}

f'(x)=1.49

Notice that the calculated slope of the tangent line is equal to the y-coordinate of the given point because f'(x)=f(x). In conclusion, the slope of the tangent line is equal to 1.49.

8 0
3 years ago
Write the polynomial f(x)=x^4-10x^3+25x^2-40x+84. In factored form
Verizon [17]
<h2>Steps:</h2>

So firstly, to factor this we need to first find the potential roots of this polynomial. To find it, the equation is \pm \frac{p}{q}, with p = the factors of the constant and q = the factors of the leading coefficient. In this case:

\textsf{leading coefficient = 1, constant = 84}\\\\p=1,2,3,4,6,7,12,14,21,28,42,84\\q=1\\\\\pm \frac{1,2,3,4,6,7,12,14,21,28,42,84}{1}\\\\\textsf{Potential roots =}\pm 1, \pm 2,\pm 3,\pm 4,\pm 6, \pm 7,\pm 12,\pm 14,\pm 21,\pm 28,\pm 42,\pm 84

Next, plug in the potential roots into x of the equation until one of them ends with a result of 0:

f(1)=(1)^4-10(1)^3+25(1)^2-40(1)+84\\f(1)=1-10+25-40+84\\f(1)=60\ \textsf{Not a root}\\\\f(2)=2^4-10(2)^3+25(2)^2-40(2)+84\\f(2)=16-10*8+25*4-80+84\\f(2)=16-80+100-80+84\\f(2)=80\ \textsf{Not a root}\\\\f(3)=3^4-10(3)^3+25(3)^2-40(3)+84\\f(3)=81-10*27+25*9-120+84\\f(3)=81-270+225-120+84\\f(3)=0\ \textsf{Is a root}

Since we know that 3 is a root, this means that one of the factors is (x - 3). Now that we know one of the roots, we are going to use synthetic division to divide the polynomial. To set it up, place the root of the divisor, in this case 3 from x - 3, on the left side and the coefficients of the original polynomial on the right side as such:

  • 3 | 1 - 10 + 25 - 40 + 84
  • _________________

Firstly, drop the 1:

  • 3 | 1 - 10 + 25 - 40 + 84
  •     ↓
  • _________________
  •     1

Next, multiply 3 and 1, then add the product with -10:

  • 3 | 1 - 10 + 25 - 40 + 84
  •     ↓ + 3
  • _________________
  •     1  - 7

Next, multiply 3 and -7, then add the product with 25:

  • 3 | 1 - 10 + 25 - 40 + 84
  •     ↓ + 3  - 21
  • _________________
  •     1  - 7 + 4

Next, multiply 3 and 4, then add the product with -40:

  • 3 | 1 - 10 + 25 - 40 + 84
  •     ↓ + 3  - 21 + 12
  • _________________
  •     1  - 7  +  4  - 28

Lastly, multiply -28 and 3, then add the product with 84:

  • 3 | 1 - 10 + 25 - 40 + 84
  •     ↓ + 3  - 21 + 12  - 84
  • _________________
  •     1  - 7  +  4  - 28 + 0

Now our synthetic division is complete. Now since the degree of the original polynomial is 4, this means our quotient has a degree of 3 and follows the format ax^3+bx^2+cx+d . In this case, our quotient is x^3-7x^2+4x-28 .

So right now, our equation looks like this:

f(x)=(x-3)(x^3-7x^2+4x-28)

However, our second factor can be further simplified. For the second factor, I will be factoring by grouping. So factor x³ - 7x² and 4x - 28 separately. Make sure that they have the same quantity inside the parentheses:

f(x)=(x-3)(x^2(x-7)+4(x-7))

Now it can be rewritten as:

f(x)=(x-3)(x^2+4)(x-7)

<h2>Answer:</h2>

Since the polynomial cannot be further simplified, your answer is:

f(x)=(x-3)(x^2+4)(x-7)

6 0
3 years ago
Which equation has the solution set x = (23) ?
Svet_ta [14]

Answer:

If you mean 2 and 3 it would be (x-2)(x-3)=0

Step-by-step explanation:

(x-2)=0

x-2=0 move the two on the other side to get positive

x=2

(x-3)=0

x-3=0 move the three on the other side to get positive

x=3

4 0
3 years ago
Can you help with a math problem
qaws [65]
I believe that they are both the first answer because when the sign has a line under it, it is darken in the circle
5 0
3 years ago
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