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3 years ago

Find the area of the squared shaped traffic sign below that is requrieed to make this square-shaped traffic sign

Mathematics

1 answer:

s2008m [1.1K]3 years ago

3 0

is a square, so it has four equal sides, so the width is 5x + 2 long and the lenght is also 5x + 2 long, so length*width, namely the area will be,

$\bf A=(5x+2)(5x+2)\implies A=\stackrel{FOIL}{25x^2+20x+4}$

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PLEASE HELP!!! THIS IS MY LAST QUESTION!!!! PLEASE!

sladkih [1.3K]

Answer:

2, -2, 0

hope it's right :)

4 0

3 years ago

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Sound policy requires that a certain number of trees be planted for every tree that is cut down for example of 8 trees are cut d

Nimfa-mama [501]

Answer:

30 trees

Step-by-step explanation:

Given the policy:

8 trees are cut down 48 trees will be planted

If homeowner is going to cut down five trees on his property how many trees will be planted

Since :

8 trees cut down = 48 planted

5 trees cut down = X

Cross multiply :

8 * X = 48 * 5

8X = 240

X = 240/8

X = 30

Hence, 30 trees will need to be planted

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3 years ago

henley needs to calculate the base of a trapezoid, b, given the area, a, in inches. She need to rearrange the formula a=16(b+4).

yKpoI14uk [10]

Answer:

b = (a/16) - 4

Step-by-step explanation:

a = 16(b + 4)

a = 16b + 64

16b + 64 = a

16b = a - 64

16b/16 = (a/16) - 64/16

b = (a/16) - 4

3 0

2 years ago

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Required information Skip to question A die (six faces) has the number 1 painted on two of its faces, the number 2 painted on th

grigory [225]

Answer:

The change to the face 3 affects the value of P(Odd Number)

Step-by-step explanation:

Analysing the question one statement at a time.

Before the face with 3 is loaded to be twice likely to come up.

The sample space is:

$S = \{1,1,2,2,2,3\}$

And the probability of each is:

$P(1) = \frac{n(1)}{n(s)}$

$P(1) = \frac{2}{6}$

$P(1) = \frac{1}{3}$

$P(2) = \frac{n(2)}{n(s)}$

$P(2) = \frac{3}{6}$

$P(2) = \frac{1}{2}$

$P(3) = \frac{n(3)}{n(s)}$

$P(3) = \frac{1}{6}$

P(Odd Number) is then calculated as:

$P(Odd\ Number) = P(1) + P(3)$

$P(Odd\ Number) = \frac{1}{3} + \frac{1}{6}$

Take LCM

$P(Odd\ Number) = \frac{2+1}{6}$

$P(Odd\ Number) = \frac{3}{6}$

$P(Odd\ Number) = \frac{1}{2}$

After the face with 3 is loaded to be twice likely to come up.

The sample space becomes:

$S = \{1,1,2,2,2,3,3\}$

The probability of each is:

$P(1) = \frac{n(1)}{n(s)}$

$P(1) = \frac{2}{7}$

$P(2) = \frac{n(2)}{n(s)}$

$P(2) = \frac{3}{7}$

$P(3) = \frac{n(3)}{n(s)}$

$P(3) = \frac{1}{7}$

$P(Odd\ Number) = P(1) + P(3)$

$P(Odd\ Number) = \frac{2}{7} + \frac{1}{7}$

Take LCM

$P(Odd\ Number) = \frac{2+1}{7}$

$P(Odd\ Number) = \frac{3}{7}$

Comparing P(Odd Number) before and after

$P(Odd\ Number) = \frac{1}{2}$ --- Before

$P(Odd\ Number) = \frac{3}{7}$ --- After

<em>We can conclude that the change to the face 3 affects the value of P(Odd Number)</em>

7 0

3 years ago

Which does not show a direct variation between x and y?

Zepler [3.9K]

A direct variation is a mathematical relationship between two variables that can be expressed by an equation in which one variable is equal to a constant times the other. In other words, a direct variation is where $y = x * (constant)$ .

In answer a, $y = x*.5$ , so it is a direct variation.
In answer c, $y = x* \frac{1}{9}$ , so it is a direct variation.
In answer d, $y = x*2$ , so it is a direct variation,

Only answer b is left, which means the answer must be 'b'.

We also know 'b' is the answer because it cannot be expressed as <span> $y = x * (constant)$ . Instead, it is expressed as </span> $y = \frac{constant}{x}$ , which is not the same thing and is therefore not a direct variation.

Hope I helped, and let me know if you have any questions :)

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3 years ago

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