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3 years ago

9

What is the value of f(−1)f(−1) when f(x)=2x 2 ? enter your answer in the box. f(−1)=

1 answer:

sergiy2304 [10]3 years ago

6 0

I hope this helps you

x= -1

f (-1)= 2. (-1)^2

f (-1)= 2.1

f (-1)= 2

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Law Incorporation [45]

The answer to this is no solution because it is in simplest form.

3x²-x+1 and 2x²+5x-4 can not be factored anymore. The expression given was in simplest form and could not be factored anymore. A trinomial expression is the form of ax²+bx+c. Factoring trinomials help you to put the equation in simplest form.

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2 years ago

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Find the area of the following figure below

icang [17]

140
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3 years ago

you get a 1 dollar coupon for every 3 salads you buy. What is the least number of salads you could buy to get 10 dollars in coup

miskamm [114]

Answer:

30 salads.

Step-by-step explanation:

If you can get a dollar coupon for 3 salads

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6 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

3. What three numbers are the Pythagorean triple generated by using 4 for x and 1 for y. Remember to use the formulas:

Gnom [1K]

Given:

The three equations for the three numbers are

$a=x^2-y^2$

$b=2xy$

$c=x^2+y^2$

To find:

The Pythagorean triple generated by using 4 for x and 1 for y.

Solution:

Substituting x=4 and and y=1 in the given equations, we get

$a=4^2-1^2$

$a=16-1$

$a=15$

In the same way find b and c.

$b=2(4)(1)$

$b=8$

The third number is

$c=4^2+1^2$

$c=16+1$

$c=17$

The required Pythagorean triple is 8, 15, and 17.

Therefore, the correct option is C.

5 0

2 years ago