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3 years ago

If lim x-> 4 f(x)=5 lim x-> 4 f g(x)=0 and lim x-> 4 f h(x)=-2, then find lim (fg)(x)

Mathematics

1 answer:

Lyrx [107]3 years ago

5 0

Answer:

Step-by-step explanation:

Assuming the problem is:

"lim x-> 4 f(x)=5 lim x-> 4 g(x)=0 and lim x-> 4 h(x)=-2, then find lim x->4 (fg)(x)"

lim x->4 (fg)(x)

Since we know the limits of f and g at x=4 exist we can write the limit as:

lim x->4 f(x) lim x->4 g(x) (since fg(x) means f times g of x.)

5(0)

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<img src="https://tex.z-dn.net/?f=0.2%2810-5c%29%20%3D%205c%20-%2016" id="TexFormula1" title="0.2(10-5c) = 5c - 16" alt="0.2(10-

san4es73 [151]

We are given,

$\longrightarrow 0.2(10 - 5c) = 5c - 16$

Solving the parenthesis

${ \longrightarrow 0.2 \times 10 - 0.2 \times 5c = 5c - 16 }$

${ \longrightarrow 2 - 1c = 5c - 16 }$

${ \longrightarrow 2 + 16 = 5c + 1c }$

${ \longrightarrow 18 = 6c }$

Dividing both sides by 6

${ \longrightarrow 3 = c }$

Hence c = 3.

8 0

3 years ago

A function g(x) is defined as shown.

FromTheMoon [43]

Answer:

g(4) = 12

Step-by-step explanation:

We need to find the value the function g(x) when x = 4, that is;

g(4)

In the first definition of the function, the value of x is strictly less than 4. Consequently, we shall use the second definition of the function to evaluate g(4)

In the second definition, g(x) is given as;

Q(x) = 0.5x + 10

plug in x = 4 and simplify;

Q(4) = 0.5(4) + 10

Q(4) = 12

6 0

3 years ago

Find f'(0.3) for f(x)= the integral from 0 to x of arccos(t)dt

Lady_Fox [76]

$\bf \textit{using the 2nd fundamental theorem of calculus}\\\\
\cfrac{dy}{dx}\displaystyle \left[ \int\limits_{0}^{x}\ cos^{-1}(t)dt \right]\implies cos^{-1}(x)
\\\\\\
f'(0.3)\iff cos^{-1}(0.3)\approx 1.26610367277949911126$

now.. 0.3 is just a value...we'e assuming Radians for the inverse cosine, so, if you check, make sure your calculator is in Radian mode

8 0

3 years ago

No links or wrong answers please!! 55 points

juin [17]

The areas of the figures are 4(x + 1), 7(d + 4) and y(y + 3)

<h3>How to determine the total areas?</h3>

<u>The figure 1</u>

In this figure, we have

Length = x + 1

Width = 4

The area is calculated as:

Area = Length * Width

So, we have

Area = 4(x + 1)

<u>The figure 2</u>

In this figure, we have

Length = d + 4

Width = 7

The area is calculated as:

Area = Length * Width

So, we have

Area = 7(d + 4)

<u>The figure 3</u>

In this figure, we have

Length = y + 3

Width = y

The area is calculated as:

Area = Length * Width

So, we have

Area = y(y + 3)

Hence, the areas of the figures are 4(x + 1), 7(d + 4) and y(y + 3)