All categories

3 years ago

Help me please. I will mark brainliest

Mathematics

1 answer:

oee [108]3 years ago

6 0

Answer:

Well, parallel lines have the same slope so I think the second option is the answer

You might be interested in

6. The graph of f(x)=-x2 + 4x – 1 is

dexar [7]

Answer:

−

1Step-by-step explanation:

sorry Just the answer

6 0

3 years ago

Determine if the functions is one to one -yes -no -see attachment-

Alex17521 [72]

Yes the functions is one to one

6 0

4 years ago

Can any one answer plz ? Is this correct

Sedbober [7]

You did that correct. Well done :)

7 0

3 years ago

Read 2 more answers

What is the following quotient? sqrt96/sqrt8

SIZIF [17.4K]

Sqrt of 96 is 9.8 and the sqrt of 8 is 2.8

9.8 divided by 2.8 is 3.5

If you don't want to solve it using decimals then:

Sqrt of 96 is 9 and the sqrt of 8 is 2

9 divided by 2 is 4.5

If you want to estimate it then:

Sqrt of 96 is 10 and the sqrt of 8 is 3

10 divided by 3 is 3.3

It all depends on whether the question wants it exact to the decimal, to the 1st number place or estimated but I think the first option of 3.5 is your best bet.

Hope that helped :-)

5 0

3 years ago

Read 2 more answers

Please me with this!

aleksandr82 [10.1K]

$sin^{-1}(x) + cos^{-1}(-\frac{1}{2}) = \pi$
$sin[(sin^{-1}(x) + cos^{-1}(-\frac{1}{2})] = 0$
$sin(sin^{-1}(x)) \cdot cos(cos^{-1}(-\frac{1}{2})) + cos(sin^{-1}(x)) \cdot sin(cos^{-1}(-\frac{1}{2})) = 0$

$-\frac{x}{2} + \sqrt{1 - x^{2}} \cdot \sqrt{1 - (-\frac{1}{2})^{2}} = 0$
$\sqrt{1 - x^{2}} \cdot \sqrt{1 - (\frac{1}{4})} = \frac{x}{2}$
$\sqrt{1 - x^{2}} \cdot \sqrt{\frac{3}{4}} = \frac{x}{2}$

$\frac{\sqrt{3(1 - x^{2})}}{2} = \frac{x}{2}$
$\sqrt{3(1 - x^{2})} = x$
$3(1 - x^{2}) = x^{2}$
$3 - 3x^{2} = x^{2}$
$4x^{2} = 3$

$x^{2} = \frac{3}{4}$
$x = \pm \frac{\sqrt{3}}{2}$

Substitute both x-values in, and only $x = \frac{\sqrt{3}}{2}$ works.

Thus, $x = \frac{\sqrt{3}}{2}$ is your solution.

7 0

3 years ago