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3 years ago

What is the maximum value for the function on the interval [-2,3]

Mathematics

2 answers:

sineoko [7]3 years ago

6 0

Answer. Second option: 2

Solution: The maximum value for the function on the interval [-2,3] is at x=-1 where the value of the function is 2 (local maximum)

Mrac [35]3 years ago

3 0

Answer: B which is 2 happy helping :)

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WILL DO ABSOLUTELY ANYTHING FOR THIS ONE HELP ASAP TIMES

Rina8888 [55]

Given:

The graph of a function $y=h(x)$ .

To find:

The interval where $h(x)>0$ .

Solution:

From the given graph graph it is clear that, the function before x=0 and after x=3.6 lies above the x-axis. So, $h(x)>0$ for $x$ and $x>3.6$ .

The function between x=0 and x=3.6 lies below the x-axis. So, $h(x)$ for $0$ .

Now,

For $-1$ , the graph of h(x) is above the x-axis. So, $h(x)>0$ .

For $0$ , the graph of h(x) is below the x-axis. So, $h(x)$ .

For $1$ , the graph of h(x) is below the x-axis. So, $h(x)$ .

Only for the interval $-1$ , we get $h(x)>0$ .

Therefore, the correct option is A.

3 0

2 years ago

A boxer is put on a diet to gain 20 lb in four weeks. The boxer gains 4 1/3 lb the first week and 5 5/6 lb the second week. How

V125BC [204]

Answer:

9.8 lbs

Step-by-step explanation:

4 1/3 + 5 5/6= 10 1/6

20 - 10 1/6 = 9.8

7 0

2 years ago

The accompanying data contains the depth (in kilometers) and magnitude, measured using the Richter Scale, of all earthquakes

Sunny_sXe [5.5K]

Answer:

Depth:

μ =20.2025 km

M = 15.625 km

Range = 47.15 km

σ ≈ 15.92 km

Q₁ = 5.7375 km

Q₃ = 34.6675 km

Magnitude:

μ = 2.08375

M = 1.465

Range, R = 5.17

σ = 1.801485 ≈ 1.8

Q₁ = 0.5625

Q₃ = 3.925

Step-by-step explanation:

The given data are;

Depth ${}$ Magnitude

0.76 ${}$ 0.84

4.93 ${}$ 0.47

8.16 ${}$ 0.35

33.58 ${}$ 1.32

21.2 ${}$ 1.61

35.03 ${}$ 4.57

10.05 ${}$ 5.52

47.91 ${}$ 1.99

For the Depth, we have;

The mean, μ = (0.76+4.93+8.16+33.58+21.2+35.03+10.05+47.91)/8 =20.2025 km

The median, M = The (n + 1)/2th term after arranging the term in increasing order as follows;

0.76, 4.93, 8.16, 10.05, 21.2, 33.58, 35.03, 47.91 , the median is therefore;

(8 + 1)/2th term or the 4.5th term which is 10.05 + (21.2 - 10.05)/2 = 15.625 km

The Range = The highest - The lowest value = 47.91 - 0.76 = 47.15 km

The Standard deviation of, σ, is given as follows;

$\sigma =\sqrt{\dfrac{\sum \left (x_i-\mu \right )^{2} }{N}}$

Where;

$x_i$ = The individual data point = (0.76, 4.93, 8.16, 10.05, 21.2, 33.58, 35.03, 47.91 )

N = The total number of data point = 8

Substituting, (using Microsoft Excel) we get;

$\sigma =\sqrt{\dfrac{\sum \left (x_i-20.2025 \right )^{2} }{8}} \approx 15.92 \ km$

Q₁ = The first quartile = The (n + 1)/4th = term arranged in increasing order

Q₁ = The (8 + 1)/4th term = The 2.25th term = 4.93 + (8.16 - 4.93)×0.25) = 5.7375 km

Q₃ = The first quartile = The 3×(n + 1)/4th = term arranged in increasing order

Q₃ = The 3×(8 + 1)/4th term = The 6.75th term = 33.58 + 3×(35.03 - 33.58)×0.25) = 34.6675 km

For the magnitude, we have, using the same formulas and procedures as above;

μ = 2.08375

M = 1.465

Range, R = 5.17

σ = 1.801485 ≈ 1.8

Q₁ = 0.5625

Q₃ = 3.925

4 0

3 years ago

There are 3 seats in each row in a school auditorium. There are 15 rows in the auditorium. Each new seat costs $74. What is the

Akimi4 [234]

3 seats x 15 rows = 45 total seats

45 seats x $74 = $3,330.00 total cost

6 0

2 years ago

How would you describe the difference between the graphs of f(x) = x^2 +4 and<br> g(y) - y^2 +4?

s2008m [1.1K]

Answer:

Step-by-step explanation:

The function $f(x)=x^2+4$ is a positive upwards opening parabola with the vertex at (0, 4), whereas

the function $g(y)=y^2+4$ is a positive rightwards opening parabola (sideways parabola) with the vertex at (4, 0). This means that answer to this is that g(y) is reflected over the x axis whereas f(x) is reflected over the y axis.

3 0

2 years ago