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3 years ago

(-1,4) and (0, -1) write a linear equation

Mathematics

1 answer:

aniked [119]3 years ago

4 0

The answer and work is shown by the picture given.

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What is the solution set of |-x|=-10?

Airida [17]

There is no solution. Because the absolute value function finds the distance between the number and 0. And length is never negative.

So, |-x| = -10 has no solution.

5 0

3 years ago

Solve C = 2πr (solve for "r")<br> C = 15

Jobisdone [24]

C = 2 pi r

but c = 15

15 = 2 pi r
15 = 2 * (22/7) * r
r = (15*7) / 2 / 22
r = 2.386

6 0

3 years ago

Read 2 more answers

The sum of two consecutive integers is − 17 -17. find the numbers

UkoKoshka [18]

I think it might be -76

4 0

3 years ago

A two-factor study with two levels of factor A and three levels of factor B uses a separate group of n = 5 participants in each

Mashcka [7]

Answer: There are 30 participants are needed for the entire study.

Step-by-step explanation:

Since we have given that

Number of levels of factor A = 2

Number of levels of factor B = 3

Number of participants in each treatment condition = 5

So, the number of participants are needed for the entire study is given by

$2\times 3\times 5\\\\=6\times 5\\\\=30$

Hence, there are 30 participants are needed for the entire study.

7 0

3 years ago

If α and β are the zeros of the quadratic polynomial f(x) = 3x2–4x + 5, find a polynomialwhose zeros are 2α + 3β and 3α + 2β.

Lelechka [254]

Answer:

$\boxed{\sf \ \ \ 3x^2-20x+37\ \ \ }$

Step-by-step explanation:

Hello,

a and b are the zeros, we can say that

$f(x)=3(x^2-\dfrac{4}{3}x+\dfrac{5}{3}) = 3(x-a)(x-b)=3(x-(a+b)x+ab)$

So we can say that

$a+b=\dfrac{4}{3}\\ab=\dfrac{5}{3}$

Now, we are looking for a polynomial where zeros are 2a+3b and 3a+2b

for instance we can write

$(x-2a-3b)(x-3a-2b)=x^2-(2a+3b+3a+2b)x+(2a+3b)(3a+2b)\\= x^2-5(a+b)x+6a^2+6b^2+9ab+4ab$

and we can notice that

$a^2+b^2=(a+b)^2-2ab$ so

$(x-2a-3b)(x-3a-2b)=x^2-5(a+b)x+6[(a+b)2-2ab]+13ab\\= x^2-5(a+b)x+6(a+b)^2+ab$

it comes

$x^2-5*\dfrac{4}{3}x+6(\dfrac{4}{3})^2+\dfrac{5}{3}$

multiply by 3

$3x^2-20x+2*16+5=3x^2-20x+37$

4 0

3 years ago