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2 years ago

The sum of two numbers is eight one number is four less than the other find the numbers

Mathematics

2 answers:

mash [69]2 years ago

4 0

Step-by-step explanation:

the numbers : x, y

x + y = 8

x = y - 4

we use the second equation in the first :

y - 4 + y = 8

2y - 4 = 8

2y = 12

y = 6

x = y - 4 = 6 - 4 = 2

LuckyWell [14K]2 years ago

3 0

Answer: The answer is 6 and 2

Step-by-step explanation:

X is first number, Y is second

x + x - 4 = 8

2x = 12

x = 6

Y = x - 4

Y = 6 - 4

Y = 2

x = 6

Y = 2

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Answer:

x>2

Step-by-step explanation:

5-x<2(x-3)+5

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2 years ago

A church steeple is 30 feet tall with square cross sections. The square at the base has side 3 feet, the square at the top has s

JulijaS [17]

Answer:

Volume= 540 ft^3

Step-by-step explanation:

Since volume= length times Width times height, you do 3 times 6 times 30= 540 ft^cubed.

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3 years ago

8 is to 32 as 1 is to

kati45 [8]

Answer:

1 is to 4

Step-by-step explanation:

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3 years ago

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DerKrebs [107]

Answer:

r = √( s/4π)

Step-by-step explanation:

s = 4πr²

s/4π = 4πr²/4π

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3 years ago

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Solve the following equations.<br> log2(x^2 − 16) − log^2(x − 4) = 1

Alenkasestr [34]

Answer:

$x=\frac{4*(2+e)}{e-2}$

Step-by-step explanation:

Let's rewrite the left side keeping in mind the next propierties:

$log(\frac{1}{x} )=-log(x)$

$log(x*y)=log(x)+log(y)$

Therefore:

$log(2*(x^{2} -16))+log(\frac{1}{(x-4)^{2} })=1\\ log(\frac{2*(x^{2} -16)}{(x-4)^{2}})=1$

Now, cancel logarithms by taking exp of both sides:

$e^{log(\frac{2*(x^{2} -16)}{(x-4)^{2}})} =e^{1} \\\frac{2*(x^{2} -16)}{(x-4)^{2}}=e$

Multiply both sides by $(x-4)^{2}$ and using distributive propierty:

$2x^{2} -32=16e-8ex+ex^{2}$

Substract $16e-8ex+ex^{2}$ from both sides and factoring:

$-(x-4)*(-8-4e-2x+ex)=0$

Multiply both sides by -1:

$(x-4)*(-8-4e-2x+ex)=0$

Split into two equations:

$x-4=0\hspace{3}or\hspace{3}-8-4e-2x+ex=0$

Solving for $x-4=0$

Add 4 to both sides:

$x=4$

Solving for $-8-4e-2x+ex=0$

Collect in terms of x and add $4e+8$ to both sides:

$x(e-2)=4e+8$

Divide both sides by e-2:

$x=\frac{4*(2+e)}{e-2}$

The solutions are:

$x=4\hspace{3}or\hspace{3}x=\frac{4*(2+e)}{e-2}$

If we evaluate x=4 in the original equation:

$log(0)-log(0)=1$

This is an absurd because log (x) is undefined for $x\leq 0$

If we evaluate $x=\frac{4*(2+e)}{e-2}$ in the original equation:

$log(2*((\frac{4e+8}{e-2})^2-16))-log((\frac{4e+8}{e-2}-4)^2)=1$

Which is correct, therefore the solution is:

$x=\frac{4*(2+e)}{e-2}$

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3 years ago