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3 years ago

Which angle is complementary to FGa

Mathematics

1 answer:

malfutka [58]3 years ago

6 0

Answer:

Is there a picture for this?

Step-by-step explanation:

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6r+7=13+7r show work please

yulyashka [42]

Answer:

r=-6

Step-by-step explanation:

6r-7r=13-7

-r=6

r=-6

8 0

3 years ago

What would the equation be for this then

Lyrx [107]

Answer:

You happen to know what I am looking at? Theres no photos :(

Step-by-step explanation:

5 0

2 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:

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

6 0

3 years ago

Can some please help me?! I would appreciate it a lot!!!

TEA [102]

hello don't worry it's easy its A

5 0

3 years ago

Parallelogram PQRS is shown on the coordinate grid. Parallelogram PQRS is dilated with the origin as the center of dilation usin

Gwar [14]

Answer:

B) Parallelogram P'Q'R'S' is smaller than parallelogram PQRS, because the scale factor is less than 1.

Step-by-step explanation:

Given that Parallelogram PQRS is shown on the coordinate grid. Parallelogram PQRS is dilated with the origin as the center of dilation using the rule (x, y) → (0.75x, 0.75y) to create parallelogram P'Q'R'S'.

This means that sides of PQRS would be 0.75 times the sides of the original PQRS

Or PQRS will be larger

B) Parallelogram P'Q'R'S' is smaller than parallelogram PQRS, because the scale factor is less than 1.

is the right answer

7 0

2 years ago