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3 years ago

1. c + 5 = 14; C = 8

Mathematics

2 answers:

Korvikt [17]3 years ago

8 0

Answer:

c can't equal 8 bc if it did, the answer would be thirteen, not fourteen. c has to equal 9.

Step-by-step explanation:

NISA [10]3 years ago

3 0

Answer:

REPORT THIS AWNSER

Step-by-step explanation:

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Solve for m and q:<br> m+ 2 = q - 4<br> 5m - 33 = q+1

NikAS [45]

Answer: m = 10 and q = 16

hope this helps!!

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3 years ago

What is the multiplicative rate of change of the exponential function shown on the graph? Two-ninths 1 4 Nine-halves

sattari [20]

Answer:

The answer is nine-halves, the last option.

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3 years ago

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

4 years ago

Plz help 14 points

ValentinkaMS [17]

1)Rewrite the table:
70, 49, 34.3, 24.01, 11.807 {The original size of the wound =70}

2) write the quotient of each number by the number before & notice the value:

49/70= 0.7
34.3/49 = 0.7
24.01/34.3 =0.7
16.0807/24.01 = 0.67 ≈0.7
You notice this is a geometric progression with r 0.7
The last term in a GP =ar^⁽ⁿ⁻¹⁾

3) Domain and Range of this function:
Last term = a₁.rⁿ⁻¹. let last term be y==> f(n) = y =70(0.7)ⁿ⁻¹
or f(n) = y = 70(0.7)ⁿ / 0.7==> f(n) = [(0.7)ⁿ ]/ 100.
This is a decreasing exponential function where the coefficient
raised to n is < 1.
The domain is for all n>= 0.
When n→∞, f(n)→0; For n=0==>f(n) =70. So the range of f(n) is:<=70

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4 years ago

The scale on a map is 1 inch : 500 miles.

Radda [10]

875 miles times 500 miles per one inch, equals 1.75 inches on the map!

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4 years ago

1. c + 5 = 14; C = 8​

1. c + 5 = 14; C = 8