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34kurt
3 years ago
6

Classify the following triangle ? check all that apply

Mathematics
1 answer:
soldier1979 [14.2K]3 years ago
7 0
What  does the triangle look like?
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The table below shows the height and the hand span of five students. What pattern of association best describes the relationship
jasenka [17]

Answer:

2

Step-by-step explanation:

2

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3 years ago
Kaleb rode his bike at a rate of 10 mph for 2 hours. How far did he ride?
Westkost [7]
20 miles! :)


Good luck! Please make me brainliest if I was correct!
7 0
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Read 2 more answers
Find the slope of the line that passes through (4,2) and (4,5)
Hunter-Best [27]

Answer: Undefined

Step-by-step explanation:

(slope = m)

m=\frac{y_2-y_1}{x_2-x_1}=\frac{5-2}{4-4}=\frac{3}{0}

(3 divided by 0 is undefined, so the slope is undefined)

4 0
3 years ago
F(x) = x2 − x − ln(x) (a) find the interval on which f is increasing. (enter your answer using interval notation.) find the inte
Mekhanik [1.2K]

Answer:

(a) Decreasing on (0, 1) and increasing on (1, ∞)

(b) Local minimum at (1, 0)

(c) No inflection point; concave up on (0, ∞)

Step-by-step explanation:

ƒ(x) = x² - x – lnx

(a) Intervals in which ƒ(x) is increasing and decreasing.

Step 1. Find the zeros of the first derivative of the function

ƒ'(x) = 2x – 1 - 1/x = 0

           2x² - x  -1 = 0

     ( x - 1) (2x + 1) = 0

         x = 1 or x = -½

We reject the negative root, because the argument of lnx cannot be negative.

There is one zero at (1, 0). This is your critical point.

Step 2. Apply the first derivative test.

Test all intervals to the left and to the right of the critical value to determine if the derivative is positive or negative.

(1) x = ½

ƒ'(½) = 2(½) - 1 - 1/(½) = 1 - 1 - 2 = -1

ƒ'(x) < 0 so the function is decreasing on (0, 1).

(2) x = 2

ƒ'(0) = 2(2) -1 – 1/2 = 4 - 1 – ½  = ⁵/₂

ƒ'(x) > 0 so the function is increasing on (1, ∞).

(b) Local extremum

ƒ(x) is decreasing when x < 1 and increasing when x >1.

Thus, (1, 0) is a local minimum, and ƒ(x) = 0 when x = 1.

(c) Inflection point

(1) Set the second derivative equal to zero

ƒ''(x) = 2 + 2/x² = 0

             x² + 2 = 0

                   x² = -2

There is no inflection point.

(2). Concavity

Apply the second derivative test on either side of the extremum.

\begin{array}{lccc}\text{Test} & x < 1 & x = 1 & x > 1\\\text{Sign of f''} & + & 0 & +\\\text{Concavity} & \text{up} & &\text{up}\\\end{array}

The function is concave up on (0, ∞).

6 0
3 years ago
Which has no real roots?
Alexxandr [17]

bruh where the problom tho

4 0
3 years ago
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