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3 years ago

Help I need help answering this question because she’s very hard can you help me3+3=

Mathematics

1 answer:

marin [14]3 years ago

6 0

Three plus three equals six :)

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Can someone help me with all 3

monitta

Answer:

1) 20

2) 60

3) 10

4) 90

Step-by-step explanation:

1. first quartile is the beginning of the box

2. third quartile is the end of the box

3. lowest is the first point of the graph

4. vice versa of 3

6 0

3 years ago

URGENT! find the sum. the answer is not D

otez555 [7]

<h3>Answer: C) -1/2</h3>

=============================================

Explanation:

We'll apply the derivative to get

g(x) = 4x^3 + 3x^2

g ' (x) = 3*4x^2 + 2*3x

g ' (x) = 12x^2 + 6x

Set that equal to 0 and solve for x. We do this to find where the horizontal tangents are located.

g ' (x) = 0

12x^2 + 6x = 0

6x(2x + 1) = 0

6x = 0 or 2x+1 = 0

x = 0 or x = -1/2

Adding those two said solutions yields us the final answer choice C

7 0

3 years ago

What happens if you label the starting number x

tester [92]

You can use any letter but when you label it x you find x

3 0

3 years ago

We are standing on the top of a 320 foot tall building and launch a small object upward. The object's vertical altitude, measure

STALIN [3.7K]

Answer:

The highest altitude that the object reaches is 576 feet.

Step-by-step explanation:

The maximum altitude reached by the object can be found by using the first and second derivatives of the given function. (First and Second Derivative Tests). Let be $h(t) = -16\cdot t^{2} + 128\cdot t + 320$ , the first and second derivatives are, respectively:

First Derivative

$h'(t) = -32\cdot t +128$

Second Derivative

$h''(t) = -32$

Then, the First and Second Derivative Test can be performed as follows. Let equalize the first derivative to zero and solve the resultant expression:

$-32\cdot t +128 = 0$

$t = \frac{128}{32}\,s$

$t = 4\,s$ (Critical value)

The second derivative of the second-order polynomial presented above is a constant function and a negative number, which means that critical values leads to an absolute maximum, that is, the highest altitude reached by the object. Then, let is evaluate the function at the critical value:

$h(4\,s) = -16\cdot (4\,s)^{2}+128\cdot (4\,s) +320$