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2 years ago

Use the graph to find the following.

Mathematics

1 answer:

Levart [38]2 years ago

5 0

a) Local maximum values of g are: 0 and 3

b) The values at which g has local maximum are : -2 and 2

<h3>What is a maximum point?</h3>

A maximum point is the point at which the gradient of the curve changes from positive to negative.

Analysis:

The inverted v-curves all have maximum values and they are two.

The one at x = -2 and the one at x = 2 for in both cases their maximum values are y = 0 and y = 3 respectively.

Learn more about minimum and maximum points: brainly.com/question/14993153

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Solve: 12.5 z – 6.4 = –8 z + 3(1.5 z – 0.5)

dolphi86 [110]

Answer:

Step-by-step explanation:

12.5z - 6.4 =-8z + 3(1.5z - 0.5)

12.5z - 6.4 = - 8z + 4.5z - 1.5

12.5z - 6.4 = - 3.5z - 1.5

Bringing like terms on one side

12.5z + 3.5z = - 1.5 + 6.4

16z = 4.9

z = 4.9/16

6 0

3 years ago

Evaluate the definite integral. <br> 1 x4(1 + 2x5)5 dx.

Nata [24]

Answer:

Step-by-step explanation:

Given the definite integral $\int\limits {\dfrac{x^4}{(1-2x^5)^5} } \, dx$ , we to evaluate it. Using integration by substitution method.

Let u = 1-2x⁵ ...1

du/dx = -10x⁴

dx = du/-10x⁴.... 2

Substitute equation 1 and 2 into the integral function and evaluate the resulting integral as shown;

$= \int\limits {\dfrac{x^4}{u^5} } \, \dfrac{du}{-10x^4}$

$= \dfrac{-1}{10} \int\limits {\dfrac{du}{u^5} } \\\\= \dfrac{-1}{10} \int\limits {{u^{-5}du } \\= \dfrac{-1}{10} [{\frac{u^{-5+1}}{-5+1}] \\\\= \dfrac{-1}{10} ({\frac{u^{-4}}{-4})\\\\$

$= \dfrac{u^{-4}}{40} \\\\\\= \dfrac{1}{40u^4} +C$

substitute u = 1-2x⁵ into the result

$= \dfrac{1}{40(1-2x^5)^4} +C$

Hence $\int\limits {\dfrac{x^4}{(1-2x^5)^5} } \, dx$ $= \dfrac{1}{40(1-2x^5)^4} +C$

5 0

3 years ago

L(t) models the length of each day (in minutes) in Manila, Philippines tt days after the spring equinox. Here, t is entered in r

SVETLANKA909090 [29]

Answer:

Given that:

$L(t) = 52\sin(\frac{2 \pi t}{365})+728$

where

L(t) represents the length of each day(in minutes) and t represents the number of days.

Substitute the value of L(t) = 750 minutes we get;

$750= 52\sin(\frac{2 \pi t}{365})+728$

Subtract 728 from both sides we get;

$22= 52\sin(\frac{2 \pi t}{365})$

Divide both sides by 52 we get;

$0.42307692352= \sin(\frac{2 \pi t}{365})$

$\frac{2 \pi t}{365} = \sin^{-1} (0.42307692352)$

Simplify:

$\frac{2 \pi t}{365} =0.43683845$

$t = \frac{365 \times 0.43683854}{2 \times \pi} = \frac{365 \times 0.43683854}{2 \times 3.14}$

Simplify:

$t \approx 25$ days

Therefore, the first day after the spring equinox that the day length is 750 minutes, is 25 days

4 0

4 years ago

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Kenya jumped 7 1/6 feet. Janet jumped 6 1/3 feet. How much farther did Kenya jump?

jolli1 [7]

Kenya jumped 5/6feet further

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3 years ago

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MZJNM = 103°. Find mZJNK<br> 2<br> *<br> 24<br> 32<br> K<br> M<br> L

just olya [345]

Answer:

m<ZJNK=47°

Step-by-step explanation:

m<ZJNK=103°-24°-32°=47°

8 0

3 years ago