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4 years ago

What are the zeros of the function? f(x) = 2x^3 – x^2 – 6x

Mathematics

1 answer:

Mars2501 [29]4 years ago

8 0

i think - 3/2?.. idk

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(06.01) In the below system, solve for y in the first equation. x + 3y = 6 2x − y = 10 one thirdx + 2 negative one thirdx + 6 −x

NISA [10]

we have that

x + 3y = 6 -------> first equation

2x − y = 10

x + 3y = 6-----> substract x both sides

-x+x+3y=6-x

3y=6-x-----> divide by 3 both sides

y=(6-x) /3

y=2-(x/3)

y=-(x/3)+2

therefore

the answer is

negative one thirdx + 2

5 0

3 years ago

Guys comment me plsssssssss.

Scrat [10]

me me me me me me me me me me me me me

8 0

3 years ago

Read 2 more answers

BEST GETS BRAINLIST! 16 POINTS! Help please, attached below, please do a step by step of The SohCahToa method. I help to find th

fenix001 [56]

Answer:

Perimeter of ΔABC = 16.46 units

Step-by-step explanation:

By applying cosine rule (CAH) in the given triangle ADB,

cos(∠BAC) = $\frac{\text{Adjacent side}}{\text{Hypotenuse}}$

cos(55°) = $\frac{3}{AB}$

AB = 5.23 units

Since, ΔABC is an isosceles triangle

AB = BC = 5.23 units

AD = DC = 3 units [Perpendicular drawn from a vertex of an isosceles triangle is a perpendicular bisector of the opposite side]

Therefore, AC = AD + DC = 6 units

Now perimeter of ΔABC = AB + BC + AC

= 5.23 + 5.23 + 6

= 16.46 units

6 0

3 years ago

I need help on this 2 problem today worth 15 points

BartSMP [9]

3. yes because -31 is lower than -27
4. it would cost $40 because you would buy 60 cups and plates

6 0

4 years ago

In 1995, the USPS approximated that they handled

STALIN [3.7K]

Answer:

$1 \times 10^{10}$ more pieces of mail were handled in 1995 than in 2010.

Step-by-step explanation:

We are given that in 1995, the USPS approximated that they handled $1.8 \times 10^{11}$ pieces of mail and in 2010, the USPS reported that they handled $1.7 \times 10^{11}$ pieces.

To find how many more pieces of mail were handled in 1995 than in 2010, we do subtraction of the pieces of mail that were handled in both the years.

Pieces of mail handled in 1995 = $1.8 \times 10^{11}$

Pieces of mail handled in 2010 = $1.7 \times 10^{11}$

As it is clear that more pieces of mail were handled in 1995.

So, Pieces of mail handled in 1995 - Pieces of mail handled in 2010 = $(1.8 \times 10^{11}) -(1.7 \times 10^{11})$

= $10^{11} \times (1.8 -1.7)$

= $10^{11} \times 0.1$ = $1 \times 10^{10}$

Hence, $1 \times 10^{10}$ more pieces of mail were handled in 1995 than in 2010.

3 0

3 years ago