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3 years ago

Let f(x)=x^2+3x−4 .

Mathematics

2 answers:

Darina [25.2K]3 years ago

6 0

Answer:

verified answer is 8

Step-by-step explanation:

Oxana [17]3 years ago

4 0

I haven't done this kind of work in a bit, but I believe it's 8.

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Find the maximum value of the function y=x^4 - x^2 +13 on the interval [-1;2]

TEA [102]

Answer: the maximum is 25.

Step-by-step explanation: a max/min can occur on the endpoints of a function and critical points of the function's derivative.

f(x)=x^4-x^2+13

f'(x)=4x^3-2x

The critical points of f'(x) occur when f'(x) is zero or undefined. f'(x) is not ever undefined in this case, so we just need to find the x values for when it's zero.

0=4x^3-2x

x=.707, -.707

Now that we have the critical points of f'(x) (.707 and -.707) and endpoints (-1 and 2), we can plug in these x values into the original function to determine its maximum. When you do this you'll find that the greatest y value produced occurs when x=2 and results in a max of 25.

4 0

3 years ago

Which could be the name of a plane?

egoroff_w [7]

Answer: where are the options?

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Solve for x<br><br> (-4/5)x + 2 =-26

ivanzaharov [21]

Answer:

(-4/5)x= -26-2

(-4/5)x= -28

x= -28/1÷ -4/5

x= -28/1 ×5/-4

x= -7/5 or -1.7

3 0

3 years ago

Someone help me pleaseeee

motikmotik

Answer:

3,943.84

Step-by-step explanation:

Since we know the angle of the arc is 70°, its length will be 7/36 of the whole circumference. The circumference, which is 2*pi*r, is 20 pi = 62.8, so the arc JL is 7/36 * 62.8 = 3,943.84

3 0

2 years ago

When fritz drives to work his trip takes 40 minutes, but when he takes the train it takes 30 minutes. find the distance fritz t

sweet-ann [11.9K]

Distance = rate*time

convert minutes to hour first because the question talking about 15 mile per hour
40 mins = 40/60 2/3 hrs
30 mins = 30/60 = 1/2hrs

Assume that s be the speed when Fritz driving, so
s + 15 will be the speed of the train.

We know the time we know the speed, Next
distance that Fritz drive = $\frac{2}{3}s$
distance the train travel = $\frac{1}{2}(s+15)$

The question: Assume that the train travels the same distance as the car
==> $\frac{2}{3}s = \frac{1}{2}(s+15)$
==> $\frac{2}{3}s = \frac{1}{2}s + \frac{15}{2}$
==> $\frac{2}{3}s - \frac{1}{2}s = \frac{15}{2}$
==> $\frac{1}{6}s = \frac{15}{2}$
==> $\frac{6}{1} * \frac{1}{6}s = \frac{15}{2} * \frac{6}{1}$
==> $s = 45$
Now we know that Fritz drive at 45 mph,
distance = $\frac{2}{3} * 45 = 30 miles$

3 0

4 years ago