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3 years ago

11

Harmons and Smiths are having a sale on apples. Harmons charges $1.25 for 5 apples. Smiths charges $2.04 for 6 apples. Find the

difference in the unit pricing.

1 answer:

8090 [49]3 years ago

7 0

Answer:

0.09

Step-by-step explanation:

Divide 1.25 by 5 which is 0.25, and 2.04 by 6 which is 0.34

Then subtract the answers 0.34-0.25=0.09

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Do yall think I can make it to a hundred on this

love history [14]

Answer:

YES

Step-by-step explanation:

because you are da best

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2 years ago

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To find 45% of 72, multiply 45/100×72.

dedylja [7]

You can also just multiply by the decimal equivalent.

12*.35=4.2

35% of 12 is 4.2

Hope this helps!

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3 years ago

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A graphing calculator is recommended. A function is given. g(x) = x4 − 5x3 − 14x2 (a) Find all the local maximum and minimum val

Taya2010 [7]

Answer:

The local maximum and minimum values are:

Local maximum

$g(0) = 0$

Local minima

$g(5.118) = -350.90$

$g(-1.368) = -9.90$

Step-by-step explanation:

Let be $g(x) = x^{4}-5\cdot x^{3}-14\cdot x^{2}$ . The determination of maxima and minima is done by using the First and Second Derivatives of the Function (First and Second Derivative Tests). First, the function can be rewritten algebraically as follows:

$g(x) = x^{2}\cdot (x^{2}-5\cdot x -14)$

Then, first and second derivatives of the function are, respectively:

First derivative

$g'(x) = 2\cdot x \cdot (x^{2}-5\cdot x -14) + x^{2}\cdot (2\cdot x -5)$

$g'(x) = 2\cdot x^{3}-10\cdot x^{2}-28\cdot x +2\cdot x^{3}-5\cdot x^{2}$

$g'(x) = 4\cdot x^{3}-15\cdot x^{2}-28\cdot x$

$g'(x) = x\cdot (4\cdot x^{2}-15\cdot x -28)$

Second derivative

$g''(x) = 12\cdot x^{2}-30\cdot x -28$

Now, let equalize the first derivative to solve and solve the resulting equation:

$x\cdot (4\cdot x^{2}-15\cdot x -28) = 0$

The second-order polynomial is now transform into a product of binomials with the help of factorization methods or by General Quadratic Formula. That is:

$x\cdot (x-5.118)\cdot (x+1.368) = 0$

The critical points are 0, 5.118 and -1.368.

Each critical point is evaluated at the second derivative expression:

x = 0

$g''(0) = 12\cdot (0)^{2}-30\cdot (0) -28$

$g''(0) = -28$

This value leads to a local maximum.

x = 5.118

$g''(5.118) = 12\cdot (5.118)^{2}-30\cdot (5.118) -28$

$g''(5.118) = 132.787$

This value leads to a local minimum.

x = -1.368

$g''(-1.368) = 12\cdot (-1.368)^{2}-30\cdot (-1.368) -28$

$g''(-1.368) = 35.497$

This value leads to a local minimum.

Therefore, the local maximum and minimum values are:

Local maximum

$g(0) = (0)^{4}-5\cdot (0)^{3}-14\cdot (0)^{2}$

$g(0) = 0$

Local minima

$g(5.118) = (5.118)^{4}-5\cdot (5.118)^{3}-14\cdot (5.118)^{2}$

$g(5.118) = -350.90$

$g(-1.368) = (-1.368)^{4}-5\cdot (-1.368)^{3}-14\cdot (-1.368)^{2}$

$g(-1.368) = -9.90$

7 0

3 years ago

Leona [35]

Answer:

309

Step-by-step explanation:

3 0

3 years ago

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I need help please.....

arlik [135]

Answer:

I am pretty sure the answer is 498.

Step-by-step explanation:

You have to solve the parentheses first. So,

9(7 (14-6) -3 x 2)

63 (8) -6)

63(8)-6

63(8) = 504

504-6 = 498

please let me know f this is correct :)

4 0

2 years ago

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