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Inga [223]
3 years ago
8

Can someone please give me a helping hand and help me answer this question

Mathematics
1 answer:
valkas [14]3 years ago
4 0

Answer:

  see attached

Step-by-step explanation:

The chart shows you that w=2 when z=0. That's the point on the w-axis at lower left. Only one equation gives those results.

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F(x)=19x+11 find f(4)​
Marysya12 [62]

Answer:

87

Step-by-step explanation:

f(x)=19x+11

Let x = 4

f(4) = 19*4 +11

     =76 +11

     = 87

3 0
3 years ago
Read 2 more answers
Use f’( x ) = lim With h ---> 0 [f( x + h ) - f ( x )]/h to find the derivative at x for the given function. 5-x²
beks73 [17]
<h2>Answer:</h2>

The derivative of the function f(x) is:

                 f'(x)=-2x

<h2>Step-by-step explanation:</h2>

We are given a function f(x) as:

f(x)=5-x^2

We have:

f(x+h)=5-(x+h)^2\\\\i.e.\\\\f(x+h)=5-(x^2+h^2+2xh)

( Since,

(a+b)^2=a^2+b^2+2ab )

Hence, we get:

f(x+h)=5-x^2-h^2-2xh

Also, by using the definition of f'(x) i.e.

f'(x)= \lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h}

Hence, on putting the value in the formula:

f'(x)= \lim_{h \to 0} \dfrac{5-x^2-h^2-2xh-(5-x^2)}{h}\\\\\\f'(x)=\lim_{h \to 0} \dfrac{5-x^2-h^2-2xh-5+x^2}{h}\\\\i.e.\\\\f'(x)=\lim_{h \to 0} \dfrac{-h^2-2xh}{h}\\\\f'(x)=\lim_{h \to 0} \dfrac{-h^2}{h}+\dfrac{-2xh}{h}\\\\f'(x)=\lim_{h \to 0} -h-2x\\\\i.e.\ on\ putting\ the\ limit\ we\ obtain:\\\\f'(x)=-2x

      Hence, the derivative of the function f(x) is:

          f'(x)=-2x

3 0
3 years ago
Read 2 more answers
The grades on a physics midterm at Covington are roughly symmetric with = 72 and = 2.0. Stephanie scored 74 on the exam. Find th
Aleksandr [31]

Answer:1.00

Step-by-step explanation:

A=74-72/2.0

And I did it on kahn

4 0
3 years ago
Read 2 more answers
2x - 5y + 3z 5x + 9y - 2z
SIZIF [17.4K]

Answer:

3xz5+2x+4y−2z

Step-by-step explanation:

2x−5y+3z5x+9y−2z

=2x+−5y+3xz5+9y+−2z

Combine Like Terms:

=2x+−5y+3xz5+9y+−2z

=(3xz5)+(2x)+(−5y+9y)+(−2z)

=3xz5+2x+4y+−2z

7 0
3 years ago
1. Use the graph of the rational function
Mrrafil [7]

Answer:

As x \to -3^{+}, f(x) \to -\infty

Step-by-step explanation:

Given:

From the graph, we can conclude that:

The function has vertical asymptotes at x=-3\ and\ x=2

The function has horizontal asymptote at f(x)=0

Vertical asymptotes are those values of 'x' for which the functions tends towards infinity. Horizontal asymptote is the value of the function as the 'x' value tends to infinity.

Now, as x \to -3^{+} means the right hand limit of the function at x=-3

From the graph, the right hand limit is the right side of the asymptote of the function at x = -3. The right side shows that the function is tending towards negative infinity.

Therefore, As x \to -3^{+}, f(x) \to -\infty

4 0
4 years ago
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