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3 years ago

Can someone solve 6÷2(1+2)

Mathematics

1 answer:

mamaluj [8]3 years ago

8 0

Answer: 9

Step-by-step explanation:

To solve this problem, we need to use our order of operations, or PEMDAS.

Parenthesis

Exponent

Multiply

Divide

Add

Subtract

6÷2(1+2) [parenthesis]

6÷2(3) [multiply/divide from left to right]

3(3) [multiply]

Now, we know that the answer is 9.

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3 years ago

Graph - 9x + 6y = 18 using linear standard form

postnew [5]

Answer:

The graph is attached Below and the plotting is given below.

Step-by-step explanation:

Given:

-9x + 6y = 18

Solution:

To draw a line on a graph the required minimum two points but here we will have three points as point A, point B, and point C.

For point A

Put x = -4 in the given equation we get

-9×-4 + 6y = 18

6y = 18-36

∴ $y=\frac{-18}{3}\\ \\y=-3$

∴ Point A ≡ ( -4, -3 ).

For point B

Put x = -2 in the given equation we get

-9×-2 + 6y = 18

6y = 18 - 18

6y = 0

∴ $y=\frac{0}{3}\\ \\y=0$

∴ Point B ≡ ( -2, 0 ).

For point C

Put x = 0 in the given equation we get

-9×0 + 6y = 18

6y = 18

∴ $y=\frac{18}{3}\\ \\y=3$

∴ Point C ≡ ( 0, 3 ).

Now we have Point A ,B and C join it and you will have Line.

3 0

3 years ago

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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4 years ago