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choli [55]
3 years ago
7

find the derivative with respect to x of f(x)=4x , and use it to find the equation of the tangent line to y=4x at x= 5

Mathematics
1 answer:
seropon [69]3 years ago
4 0

Answer:

F(x)=4X X=5

F(5)=4X

4(5)

20

Y=4(20)

Y=80

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What is the slope of the line that contains the points (-1,-1 and (3,15?
DENIUS [597]
Slope= y2-y1 / x2-x1
slope = 15 - (-1) / 3- (-1)
slope = 16 / 4
slope = 4
7 0
3 years ago
The University of Central Florida's cheerleading team has eighteen males and twenty-one females. If h represents the height of a
Anuta_ua [19.1K]

Answer:

The range of heights of the cheerleaders is the interval [58, 74)

All real numbers greater than or equal to 58 inches and less than 74 inches

Step-by-step explanation:

we have

260 \leq 4h+28

Divide the compound inequality into two inequalities

260 \leq 4h+28 -----> inequality A

4h+28 -----> inequality B

Solve inequality A

260 \leq 4h+28

Subtract 28 both sides

232 \leq 4h

Divide by 4 both sides

58 \leq h

Rewrite

h \geq 58\ in

Solve the inequality B

4h+28

Subtract 28 both sides

4h

Divide by 4 both sides

h

therefore

The range of heights of the cheerleaders is the interval [58, 74)

All real numbers greater than or equal to 58 inches and less than 74 inches

3 0
3 years ago
24 is 4 times as great as<br> Pls help
Vaselesa [24]

Answer:

6=k

Step-by-step explanation:

5 0
3 years ago
A soccer ball is kicked from 3 feet above the ground. the height,h, in feet of the ball after t seconds is given by the function
dexar [7]
Here's our equation.

h=-16t+64t+3

We want to find out when it returns to ground level (h = 0)

To find this out, we can plug in 0 and solve for t.

0 = -16t+64t+3 \\ 16t-64t-3=0 \\ use\ the\ quadratic\ formula\ \frac{-b\±\sqrt{b^2-4ac}}{2a}  \\ \frac{-(-64)\±\sqrt{(-64)^2-4(16)(-3)}}{2*16}  = \frac{64\±\sqrt{4096+192}}{32}

= \frac{64\±\sqrt{4288}}{32} = \frac{64\±8\sqrt{67}}{32} = \frac{8\±\sqrt{67}}{4} = \boxed{\frac{8+\sqrt{67}}{4}\ or\ 2-\frac{\sqrt{67}}{4}}

So the ball will return to the ground at the positive value of \boxed{\frac{8+\sqrt{67}}{4}} seconds.

What about the vertex? Simple! Since all parabolas are symmetrical, we can just take the average between our two answers from above to find t at the vertex and then plug it in to find h!

\frac{1}2(\frac{8+\sqrt{67}}{4}+2-\frac{\sqrt{67}}{4}) = \frac{1}2(2+\frac{\sqrt{67}}{4}+2-\frac{\sqrt{67}}{4}) = \frac{1}2(4) = 2

h=-16t^2+64t+3 \\ h=-16(2)^2+64(2)+3 \\ \boxed{h=67}


8 0
4 years ago
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olga2289 [7]

Answer:can’t read it

Step-by-step explanation:

Can you take a pic closer to the screen?

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3 years ago
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