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UNO [17]
3 years ago
5

Mark runs 3/4 of a mile each day for 5 days. What is the total distance that Mark has run after 5 days? A.3 3/4 B.4 1/4 C.5 3/4

D.6 2/3
Mathematics
1 answer:
Evgen [1.6K]3 years ago
6 0
3/4 x 5 = 3 3/4

Your answer is A.
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Graph triangle XYZ with vertices X(0, -4), Y (4, -1), and Z(6, -3) on the coordinate grid.
gogolik [260]

Answer:

Step-by-step explanation:

6 0
2 years ago
Does 5 go into 100 equally
andreev551 [17]
Yes, 5 goes into 100 exactly 20 times because 5 multiplied by 20 is 100
6 0
2 years ago
You throw a ball up and its height h can be tracked using the equation h=2x^2-12x+20.
postnew [5]

<em><u>This problem seems to be wrong because no minimum point was found and no point of landing exists</u></em>

Answer:

1) There is no maximum height

2) The ball will never land

Step-by-step explanation:

<u>Derivatives</u>

Sometimes we need to find the maximum or minimum value of a function in a given interval. The derivative is a very handy tool for this task. We only have to compute the first derivative f' and have it equal to 0. That will give us the critical points.

Then, compute the second derivative f'' and evaluate the critical points in there. The criteria establish that

If f''(a) is positive, then x=a is a minimum

If f''(a) is negative, then x=a is a maximum

1)

The function provided in the question is

h(x)=2x^2-12x+20

Let's find the first derivative

h'(x)=4x-12

solving h'=0:

4x-12=0

x=3

Computing h''

h''(x)=4

It means that no matter the value of x, the second derivative is always positive, so x=3 is a minimum. The function doesn't have a local maximum or the ball will never reach a maximum height

2)

To find when will the ball land, we set h=0

2x^2-12x+20=0

Simplifying by 2

x^2-6x+10=0

Completing squares

x^2-6x+9+10-9=0

Factoring and rearranging

(x-3)^2=-1

There is no real value of x to solve the above equation, so the ball will never land.

This problem seems to be wrong because no minimum point was found and no point of landing exists

3 0
3 years ago
How many 'words' can be made from the name ESTABROK with no restrictions
Bumek [7]

The number of ways in which the name 'ESTABROK' can be made with no restrictions is 40, 320 ways.

<h3>How to determine the number of ways</h3>

Given the word:

ESTABROK

Then n = 8

p = 6

The formula for permutation without restrictions

P = n! ( n - p + 1)!

P = 8! ( 8 - 6 + 1) !

P = 8! (8 - 7)!

P = 8! (1)!

P = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 × 1

P = 40, 320 ways

Thus, the number of ways in which the name 'ESTABROK' can be made with no restrictions is 40, 320 ways.

Learn more about permutation here:

brainly.com/question/4658834

#SPJ1

6 0
1 year ago
What is the area of the figure at the right?
Anvisha [2.4K]

Answer: 88 in2

Step-by-step explanation:

16 * 4 = 64

8 * 3 = 24 / 2 = 12

12 + 12 + 64 = 88

4 0
2 years ago
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