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ruslelena [56]
3 years ago
7

Kimberly wants to enlarge hut 5‘ x 10‘ rectangular patio by multiplying its dimensions by two how old the area of the patio be a

ffected
Mathematics
1 answer:
deff fn [24]3 years ago
8 0

Answer:

1/4 the new area

Step-by-step explanation:

Kimberly wants to enlarge hut 5‘ x 10‘ rectangular patio by multiplying its dimensions by two how old the area of the patio be affected

Given that:

Initial dimension of patio = 5' by 10'

Initial area of patio = Area of rectangle = Length * width

Area of patio = 5 * 10 = 50 in²

If dimension is multiplied by 2

New dimension :

5* 2 by 10*2 = 10' by 20'

New area of patio :

10 * 20 = 200 in²

50 /200 = 1/4

The old area will be 1/4 the size of the new area of the patio

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Let D = {-48, -14, -8, 0, 1, 3, 16, 23, 26, 32, 36} Determine which of the following statements are true and which are false. a)
morpeh [17]

Answer:

\forall x\in D if x is odd then x> 0 is true statement.

\forall x\in D if x is odd then x> 0 is true statement.

\forall x\in D if x is even then x≤0 is false statement.

\forall x\in D If the ones digit of x is 2, then the tens digit is 3 or 4 is true statement.

\forall x\in D if the ones digit of x is 6, then the tens digit is 1 or 2 is false statement.

Step-by-step explanation:

Consider the provided information.

D = {-48, -14, -8, 0, 1, 3, 16, 23, 26, 32, 36}

Part (A) \forall x\in D if x is odd then x> 0

Here only even numbers are less than 0 that means the statement is true.

\forall x\in D if x is odd then x> 0 is true statement.

Part (B) \forall x\in D if x is less than 0 then x is even.

Here only even numbers are less than 0 that means the statement is true.

\forall x\in D if x is odd then x> 0 is true statement.

Part (C) \forall x\in D if x is even then x≤0

Here we can see that 16, 26, 32, 36 are even number and also greater than 0. Thus the statement is false.

\forall x\in D if x is even then x≤0 is false statement.

Part (D) \forall x\in D If the ones digit of x is 2, then the tens digit is 3 or 4.

There is only one number whose ones digit is 2. i.e. 32 also the tens digit of the number 32 is 3. Which makes the above statement true.

\forall x\in D If the ones digit of x is 2, then the tens digit is 3 or 4 is true statement.

Part (E) \forall x\in D if the ones digit of x is 6, then the tens digit is 1 or 2.

Numbers having ones digit 6 are: 16, 26 and 36

Here, the tens digits are 1, 2 and 3 which is contradict to our statement. Hence the provided statement is false.

\forall x\in D if the ones digit of x is 6, then the tens digit is 1 or 2 is false statement.

8 0
3 years ago
P(x) = 3x2 - 7x + 11, find P(9).
nadya68 [22]
You need to substitute 9 with every x value in the equation. 
So, 3(9)^2-7(9)+11
p(x)= 191
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3 years ago
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Please, can you do this step-by-step.... 2 1/3 + 6 3/5
lina2011 [118]

Answer:

8 14/15

Step-by-step explanation:

Step 1:

2 1/3 + 6 3/5

Step 2:

2 5/15 + 6 9/15

Answer:

8 14/15

Hope This Helps :)

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Answer

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Step-by-step explanation:

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4 years ago
Square roots in trigonometry. I don’t understand please help?
cupoosta [38]

By definitions of the (co)tangent and cosecant function,

3\tan^2x-2=\csc^2x-\cot^2x\iff3\dfrac{\sin^2x}{\cos^2x}-2=\dfrac1{\sin^2x}-\dfrac{\cos^2x}{\sin^2x}

Turn everything into fractions with common denominators:

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Recall that \cos^2x+\sin^2x=1, so we can simplify both sides a bit.

On the left:

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On the right:

\dfrac{1-\cos^2x}{\sin^2x}=\dfrac{\sin^2x}{\sin^2x}=1

(as long as \sin x\neq 0, which happens in the interval 0\le x\le\pi when x=0 or x=\pi)

So we have

\dfrac{3-5\cos^2x}{\cos^2x}=1\implies3-5\cos^2x=\cos^2x

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\implies\cos x=\pm\dfrac1{\sqrt2}

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