Is (-1,2) a solution of y=-2x-1

2 answers:

Klio2033 [76]3 years ago

7 0

A solution of a function is where the function hits the x-axis, so if y is any number that is not equal to 0, there is no possible way that point can be a solution.

Hope this helps!

chubhunter [2.5K]3 years ago

4 0

The solution is: y=1.

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Help pls. cant figure it out.

const2013 [10]

A=2

4*2=8+1=9/////2+7=9
both of the answers are 9, so a=2

8 0

3 years ago

Read 2 more answers

Log_6⁡(x+1)+log_6⁡x=1

Fiesta28 [93]

This is pretty simple once you figure out the trick.

All you have to do is make the args common.

Meaning $log_{6}(x + 1) + log_{6}(x) = 1$ is nothing but $log_{6}(x(x + 1)) = 1$

Which is nothing but: $log_{6}(x^2 + x) = 1$ .

Now using the scorpion tail method, this equation can be reframed to being: $6 = x^2 + x$ . You bring 6 to the other side and it'll become: $x^2 + x - 6 = 0$

Factorise it and you get: $(x - 2) * (x + 3)$

Henceforth, the solution would be $2$ and $-3$ . :D

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3 years ago

Please answer correctly !!!!!! Will mark brainliest !!!!!!!!!!

KatRina [158]

Answer:

$2a^2\sqrt{35a}$

Step-by-step explanation:

$\sqrt{2a} .\sqrt{14a^3} .\sqrt{5a} \\=\sqrt{2a\times 14a^3\times 5a}\\=\sqrt{140a^5}\\=2a^2\sqrt{35a}\\$

8 0

3 years ago

A bag contains 6 red apples and 5 yellow apples. 3 apples are selected at random. Find the probability of selecting 1 red apple

tester [92]

Answer: The required probability of selecting 1 red apple and 2 yellow apples is 36.36%.

Step-by-step explanation: We are given that a bag contains 6 red apples and 5 yellow apples out of which 3 apples are selected at random.

We are to find the probability of selecting 1 red apple and 2 yellow apples.

Let S denote the sample space for selecting 3 apples from the bag and let A denote the event of selecting 1 red apple and 2 yellow apples.

Then, we have

$n(S)=^{6+5}C_3=^{11}C_3=\dfrac{11!}{3!(11-3)!}=\dfrac{11\times10\times9\times8!}{3\times2\times1\times8!}=165,\\\\\\n(A)\\\\\\=^6C_1\times^5C_2\\\\\\=\dfrac{6!}{1!(6-1)!}\times\dfrac{5!}{2!(5-2)!}\\\\\\=\dfrac{6\times5!}{1\times5!}\times\dfrac{5\times4\times3!}{2\times1\times3!}\\\\\\=6\times5\times2\\\\=60.$