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devlian [24]
3 years ago
9

in a art class, there are 48 pencils to 56 crayons. what is the ratio of pincels to total written as a fraction in simplest form

​
Mathematics
1 answer:
Anit [1.1K]3 years ago
6 0

Answer: 48:104

Step-by-step explanation: 48 is the amount of pencils 48+56=108 is the total

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Explain why the mapping diagram below represents a function.
tiny-mole [99]

Answer:

Because there is only one out put for each domain this mapping diagram is a function.

Step-by-step explanation:

Two sets are involved.  There is a directional property, the relation is defined from one set called the domain on to another set called the codomain. A function is a relation that has exactly one output for each input in the domain.

6 0
3 years ago
ABC is translated 2 units right and 1 unit down. What is the new location of point C?
Bogdan [553]

Answer:

(4,-8)

Step-by-step explanation

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8 0
3 years ago
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Find the missing side of the triangle
Katena32 [7]

\\ \sf\longmapsto sin\theta=\dfrac{Perpendicular}{Hypotenuse}

\\ \sf\longmapsto sin74=\dfrac{n}{40}

\\ \sf\longmapsto n=40sin74

\\ \sf\longmapsto n=35(0.96)

\\ \sf\longmapsto n=38.4

7 0
2 years ago
A. 115<br> B. 167<br> C. 126<br> D. 96
geniusboy [140]

Answer:

126

Step-by-step explanation:

Let x be the missing length

The triangles are similar:

● UE/140 = 45/x

From the graph we deduce that:

● UE = 140 - 90 = 50

Replace UE by its value

● 50/ 140 = 45/x

Switch x and 50

● x / 140 = 45/50

45/50 is 9/10 wich is 0.9

● x/140 = 0.9

Multiply 0.9 by 140

● x = 140 × 0.9

● x = 126

5 0
3 years ago
Read 2 more answers
F(x) = x2 − x − ln(x) (a) find the interval on which f is increasing. (enter your answer using interval notation.) find the inte
Mekhanik [1.2K]

Answer:

(a) Decreasing on (0, 1) and increasing on (1, ∞)

(b) Local minimum at (1, 0)

(c) No inflection point; concave up on (0, ∞)

Step-by-step explanation:

ƒ(x) = x² - x – lnx

(a) Intervals in which ƒ(x) is increasing and decreasing.

Step 1. Find the zeros of the first derivative of the function

ƒ'(x) = 2x – 1 - 1/x = 0

           2x² - x  -1 = 0

     ( x - 1) (2x + 1) = 0

         x = 1 or x = -½

We reject the negative root, because the argument of lnx cannot be negative.

There is one zero at (1, 0). This is your critical point.

Step 2. Apply the first derivative test.

Test all intervals to the left and to the right of the critical value to determine if the derivative is positive or negative.

(1) x = ½

ƒ'(½) = 2(½) - 1 - 1/(½) = 1 - 1 - 2 = -1

ƒ'(x) < 0 so the function is decreasing on (0, 1).

(2) x = 2

ƒ'(0) = 2(2) -1 – 1/2 = 4 - 1 – ½  = ⁵/₂

ƒ'(x) > 0 so the function is increasing on (1, ∞).

(b) Local extremum

ƒ(x) is decreasing when x < 1 and increasing when x >1.

Thus, (1, 0) is a local minimum, and ƒ(x) = 0 when x = 1.

(c) Inflection point

(1) Set the second derivative equal to zero

ƒ''(x) = 2 + 2/x² = 0

             x² + 2 = 0

                   x² = -2

There is no inflection point.

(2). Concavity

Apply the second derivative test on either side of the extremum.

\begin{array}{lccc}\text{Test} & x < 1 & x = 1 & x > 1\\\text{Sign of f''} & + & 0 & +\\\text{Concavity} & \text{up} & &\text{up}\\\end{array}

The function is concave up on (0, ∞).

6 0
3 years ago
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