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Vaselesa [24]
3 years ago
6

The amount of snow on the ground increased by 4 inches between 4 p.m. and 6 p.m. If there was 6 inches of snow on the ground at

4 p.m. how many inches were on the ground at 6 p.m?
A
10 inches

B
14 inches

C
2 inches

D
18 inches
Mathematics
2 answers:
svet-max [94.6K]3 years ago
8 0

Answer:

A.ten inches

Step-by-step explanation:

6 plus 4 =10

Neporo4naja [7]3 years ago
7 0

Answer:

A

Step-by-step explanation:

6 at  4 pm + 4 that fell between 4 and 6 pm = 10 (A)

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A bag of marbles has 5 blue marbles, 3 red marbles, and 6 yellow marbles. A marble is drawn from the bag at
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Step-by-step explanation:

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3 years ago
Read 2 more answers
In relation t a functions? Is the inverse of relation t function
frozen [14]

Answer:

Relation t is a function. The inverse of relation t is not a function ⇒ 3rd

Step-by-step explanation:

* Lets explain how to solve the problem

- A relation is a set of inputs and outputs, and a function is a relation

 with one output for each input

- Ex: T = {(1 , 2) , (3 , 5) , (-4 , 0)} is a function because every input has

  only one output

- To find the inverse of a function we switched the input and the

  out put

- The inverse of a function may not always be a function

* Lets solve the problem

∵ The relation t is the set of ordered pairs of x and y

  x = 0     2      4    6

  y = -8    -7    -4    -4

∵ x = 0 has only y = -8

∵ x = 2 has only y = -7

∵ x = 4 has only y = -4

∵ x = 6 has only y = -4

∴ Every value of x has only one value of y

∴ Relation t is a function

* Lets find its inverse by switching x and y

∵ The inverse function of t is :

   x = -8     -7     -4    -4

   y =  0      2      4     6

∵ x = -8 has only y = 0

∵ x = -7 has only y = 2

∵ x = -4 has y = 4 and y = 6

∴ Not every value of x has only one value of y

∴ The inverse of relation t is not a function

* Relation t is a function. The inverse of relation t is not a function

4 0
3 years ago
Use a linear approximation (or differentials) to estimate the given number. (Round your answer to five decimal places.) 3 217
Soloha48 [4]

Answer:

f(216) \approx 6.0093

Step-by-step explanation:

Given

\sqrt[3]{217}

Required

Solve

Linear approximated as:

f(x + \triangle x) \approx f(x) +\triangle x \cdot f'(x)

Take:

x = 216; \triangle x= 1

So:

f(x) = \sqrt[3]{x}

Substitute 216 for x

f(x) = \sqrt[3]{216}

f(x) = 6

So, we have:

f(x + \triangle x) \approx f(x) +\triangle x \cdot f'(x)

f(215 + 1) \approx 6  +1 \cdot f'(x)

f(216) \approx 6  +1 \cdot f'(x)

To calculate f'(x);

We have:

f(x) = \sqrt[3]{x}

Rewrite as:

f(x) = x^\frac{1}{3}

Differentiate

f'(x) = \frac{1}{3}x^{\frac{1}{3} - 1}

Split

f'(x) = \frac{1}{3} \cdot \frac{x^\frac{1}{3}}{x}

f'(x) = \frac{x^\frac{1}{3}}{3x}

Substitute 216 for x

f'(216) = \frac{216^\frac{1}{3}}{3*216}

f'(216) = \frac{6}{648}

f'(216) = \frac{3}{324}

So:

f(216) \approx 6  +1 \cdot f'(x)

f(216) \approx 6  +1 \cdot \frac{3}{324}

f(216) \approx 6  + \frac{3}{324}

f(216) \approx 6  + 0.0093

f(216) \approx 6.0093

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3 years ago
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marishachu [46]

Answer:

Step-by-step explanation:

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3 years ago
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