PLZZZZZZZ NEED HELP!!!!!!!!!

Leokris [45]

(a) mtan refers to the slope of the tangent line. Given f(x) = 9 + 7x ², compute the difference quotient:

$\dfrac{f(x+h)-f(x)}h = \dfrac{(9+7(x+h)^2)-(9+7x^2)}h = \dfrac{(9+7x^2+14xh+7h^2)-9-7x^2}h = \dfrac{14xh+7h^2}h$

Then as h approaches 0 - bearing in mind that we're specifically considering h near 0, and not h = 0 - we can eliminate the factor of h in the numerator and denominator, so that

$m_{\rm tan} = \displaystyle \lim_{h\to0}\frac{f(x+h)-f(x)}h = \lim_{h\to0}\frac{14xh+7h^2}h = \lim_{h\to0}(14x+7h) = 14x$

and so the slope of the line at P (0, 9), for which we take x = 0, is 0.

(b) The equation of the tangent line is then y = 9.

4 0

2 years ago

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Olegator [25]

Answer:

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8 0

3 years ago

Read 2 more answers

Adding and subtracting polynomials (-3x² - 9x + 5) - ( -2x²- 4x + 7)

seraphim [82]

Answer:

(-3x²- 9x+5) - (-2x²-4x +7)

= (-3x²- 9x²+5) (2x²+4x +7)

=-3x²-9x+5+2x²+4x-7

=-3x²+2x²-9x+4x+5-7

=-x²-5x-2

6 0

3 years ago

Eva bought popcorn, candy and a drink at the movies. The popcorn was three times as expensive as the candy and the drink was twi

aliina [53]

The cost of popcorn, drinks, and candy is 5.25, 3.5, and 1.75.

Given,

Eva bought popcorn, candy, and a drink at the movies.

The popcorn was three times as expensive as the candy and the drink was twice as expensive as the candy.

Eva spent a total of 10.50

We need to find the cost of popcorn, candy, and drink.

We have,

Let the cost of the candy be C.

Now,

The popcorn was three times as expensive as the candy.

Cost of the Popcorn = 3C

The drink was twice as expensive as the candy.

Cost of the drink = 2C

Eva spent a total of 10.50

Cost of popcorn + cost of drink + cost of candy = 10.50

3C + 2C + C = 10.50

Adding

6C = 10.50

Dividing by 6 into both sides

C = 10.50/6

C = 1.75.

Now,

Cost of the Popcorn = 3C = 3 x 1.75 = 5.25

Cost of the drink = 2C = 2 x 1.75 = 3.5

Cost of the candy = C = 1.75

Thus the cost of popcorn, drinks, and candy is 5.25, 3.5, and 1.75.

Learn more about finding the cost of items with the given total cost here:

brainly.com/question/357628

#SPJ1

6 0

2 years ago

What is the value of the interquartile range of the data below? 6 12 14 24

Strike441 [17]

Answer:

14

Step-by-step explanation:

The interquartile range is the value of quartile 3 minus quartile 1.

The "box" part of this diagram has 3 lines, the left most, the middle, and the rightmost.

The leftmost line is Quartile 1. The right most is Quartile 3.

Hence

interquartile range = quartile 3 - quartile 1

Looking at the box plot, we can see that each small line in the number line is 2 units.

The leftmost line (quartile 1 ) is at 1 unit left of 30, so that is 30 -2 = 28

The rightmost line (quartile 3) is at 1 unit right of 40, so that is 40 + 2 = 42

Hence,

Interquartile range = 42 - 28 = 14

7 0

4 years ago