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3 years ago

5(x + y) -3( x - y)=

Mathematics

2 answers:

exis [7]3 years ago

7 0

5(x + y) - 3(x - y)

5x + 5y - 3x + 3y

5x - 3x + 5y + 3y

2x + 8y

5(x + y) - 3(x - y) = 2x + 8y

faust18 [17]3 years ago

3 0

Answer:

2x+2y

Step-by-step explanation:

5(x+y) - 3(x-y)

= 5x+5y-3x-3y

=5x-3y+5y-3y

=2x+2y

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Is 0 not an element of an empty set?

kirza4 [7]

Answer:

0 is not an element of an empty set.

Step-by-step explanation:

We are asked to determine whether 0 is not an element of an empty set.

We know that an empty set is an unique set having no elements. The cardinality of an empty set is 0.

Cardinality stands for the count of element is an set. An empty set is denoted by symbols ∅ or { }.

The empty set is like an empty container. The container is there, but nothing is in it.

When 0 is an element of a set, then its cardinality would be 1.

Therefore, 0 is not an element of an empty set.

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4 years ago

Bruno uses a piece of wrapping paper with dimensions 1 and 1/4 feet by 3 feet to wrap a gift. What is the total area of the pape

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Answer: $A=3\dfrac{3}{4}\ ft^2$

Step-by-step explanation:

Given: The dimensions of the wrapping paper :

$1\text{ and}\dfrac{1}{4}\text{ feet }\text{ by 3 feet}\\\\\text{i.e }1+\dfrac{1}{4}\text{ feet }\text{ by 3 feet}\\\\\text{i.e }\dfrac{5}{4}\text{ feet }\text{ by 3 feet}$

Now, we know that the area of a rectangle is given by :-

$A=length*width$

Now, the area of the wrapping paper is given by :-

$A=\dfrac{5}{4}\times3=\dfrac{15}{4}=3\dfrac{3}{4}\ ft^2$

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3 years ago

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3 years ago

For the function f(x) = x2 -12, find (f o f-1)(4)

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3 years ago

If y = e2x is a solution to y''- 5y' + ky = 0, what is the value of k?

Nutka1998 [239]

Answer:

The value of k is 6

Step-by-step explanation:

we need to find the value of k

Given : - $y=e^{2x}$ is the solution $y''-5y'+ky=0$

$y=e^{2x}$ ........(1)

differentiate $y=e^{2x}$ with respect to 'x'

$\frac{dy}{dx}=\frac{d}{dx}e^{2x}$

Since, $\frac{d}{dx}e^{x} =e^{x}\frac{d}{dx}(x)$

$\frac{dy}{dx}=e^{2x}\frac{d}{dx}(2x)$

$\frac{dy}{dx}=e^{2x}\times 2$

$\frac{dy}{dx}=2e^{2x}$

so, $y'=2e^{2x}$ ..........(2)

Again differentiation above with respect to 'x'

$\frac{d}{dx}\frac{dy}{dx}=\frac{d}{dx}2e^{2x}$

$\frac{d^{2}y}{dx^{2}}=2e^{2x}\frac{d}{dx}(2x)$

$\frac{d^{2}y}{dx^{2}}=2e^{2x}\times 2$

$\frac{d^{2}y}{dx^{2}}=4e^{2x}$

so, $y''=4e^{2x}$ ........(3)

Now, put the value of $y\ ,y' \ \text{and} \ y''$ in $y''-5y'+ky=0$

$4e^{2x}-5(2e^{2x})+(e^{2x})k=0$

$4e^{2x}-10e^{2x}+e^{2x}k=0$

$-6e^{2x}+e^{2x}k=0$

add both the sides by $6e^{2x}$

$e^{2x}k=6e^{2x}$

Cancel out the same terms from left and right sides

$k=6$

Hence, the value of k is 6

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4 years ago