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lakkis [162]
3 years ago
8

20. Please help I really need it​

Mathematics
1 answer:
sertanlavr [38]3 years ago
5 0

Step-by-step explanation:

d if not d a.................

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What happens to the function f(x)=x4 when it becomes f′(x)=x4+5?
elena-s [515]
I think it moves 5 units to the right Im not sure tho!!
8 0
2 years ago
Graph y = x2 + 2. Identify the vertex of the graph. Tell whether it is a minimum or maximum
Leona [35]
Vertex is -b/2a. The variable b=2 and a=2. That finds the x coordinate of the vertex. Then to find the y coordinate you just plug the number you got for x (-1/2) into the equation. Like so:
y = 2x+2
y = 2(-1/2)+2
y = -1+2
y = 1
So your vertex is ( -1/2, 1 )
To graph your vertex you take your x coordinate (-1/2) and find it on the x axis of the graph (horizontal line). Then do the same for y (1) (on the vertical axis.
Now that you have your vertex graphed you can find out if it is minimum or maximum by looking at the direction in which the slope is going. If it goes in the positive direction (up) then it is a minimum since that is the lowest point that the line touches. If it is maximum it is opposite. The line faces down and the highest point the line touches is that point.
4 0
3 years ago
Identify the expression with nonnegative limit values. More info on the pic. PLEASE HELP.
marshall27 [118]

Answer:

\lim _{x\to 2}\:\frac{x-2}{x^2-2}\\\\  \lim _{x\to 11}\:\frac{x^2+6x-187}{x^2+3x-154}\\\\ \lim _{x\to \frac{5}{2}}\left\frac{2x^2+x-15}{2x-5}\right

Step-by-step explanation:

a) \lim _{x\to 3}\:\frac{x^2-10x+21}{x^2+4x-21}=\lim \:_{x\to \:3}\:\frac{\left(x-7\right)\left(x-3\right)}{\left(x+7\right)\left(x-3\right)}=\lim \:_{x\to \:3}\:\frac{x-7}{x+7}=\frac{3-7}{3+7}=-\frac{4}{10}=-\frac{2}{5}

b) \lim _{x\to -\frac{3}{2}}\left(\frac{2x^2-5x-12}{2x+3}\right)=\lim \:_{x\to -\frac{3}{2}}\:\frac{\left(2x+3\right)\left(x-4\right)}{\left(2x+3\right)}=\lim \:\:_{x\to \:-\frac{3}{2}}\:\left(x-4\right)=-\frac{3}{2}-4\\ \\ \lim _{x\to -\frac{3}{2}}\left(\frac{2x^2-5x-12}{2x+3}\right)=-\frac{11}{2}

c) \lim _{x\to 2}\:\frac{x-2}{x^2-2}=\frac{2-2}{\left(2\right)^2-2}=\frac{0}{4-2}=0

d) \lim _{x\to 11}\:\frac{x^2+6x-187}{x^2+3x-154}=\lim _{x\to 11}\:\frac{\left(x-11\right)\left(x+17\right)}{\left(x-11\right)\left(x+14\right)}=\lim _{x\to 11}\:\frac{\left(x+17\right)}{\left(x+14\right)}=\frac{11+17}{11+14}=\frac{28}{25}

e) \lim _{x\to 3}\:\frac{x^2-8x+15}{x-3}=\lim \:_{x\to \:3}\:\frac{\left(x-3\right)\left(x-5\right)}{x-3}=\lim _{x\to 3}\left(x-5\right)=3-5=-2

f) \lim _{x\to \frac{5}{2}}\left(\frac{2x^2+x-15}{2x-5}\right)=\lim \:_{x\to \:\frac{5}{2}}\frac{\left(2x-5\right)\left(x+3\right)}{2x-5}=\lim \:\:_{x\to \:\:\frac{5}{2}}\left(x+3\right)=\frac{5}{2}+3=\frac{11}{2}

4 0
3 years ago
Can someone please help me with this question
Rudiy27
A. 30.4 minutes

Hope this helps
3 0
2 years ago
Read 2 more answers
The width-to-height ratio of a certain computer monitor
Mnenie [13.5K]
<h2>24 inches</h2>

Step-by-step explanation:

It is given that \frac{\textbf{width of computer monitor}}{\textbf{height of computer monitor}}=\frac{16}{9}

It is given that \text{height of computer monitor}=13.5

\text{width of computer monitor} =\text{height of computer monitor}\times \frac{16}{9}

\text{width of computer monitor}=13.5\times \frac{16}{9} =\frac{216}{9} =24

Hence the required width in inches is 24.

5 0
3 years ago
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