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3 years ago

Please Find g(x)-h(x)

Mathematics

1 answer:

andrey2020 [161]3 years ago

5 0

Answer:

Step-by-step explanation:

g(x) - h(x)

= 5x² - 10x + 9 - ( - 3x³ + 5x - 6) ← distribute parenthesis by - 1

= 5x² - 10x + 9 + 3x³ - 5x + 6 ← collect like terms

= 3x³ + 5x² - 15x + 15 → B

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I got a lotnof missing work

SashulF [63]

Each eraser was $4.

6+4(3)=18

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please vote my answer brainliest. thanks!

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3 years ago

A particle moves along a straight line. The distance of the particle from the origin at time t is modeled by the equation below.

laiz [17]

Answer:

The value of t that will satisfy the equation is π/6 (which is 30 degrees)

Step-by-step explanation:

The function that models the movement of the particle is given as;

S(t) = 2 sin(t) + 3 cos (t)

Now we want to the value of t between 0 and pi/2 that satisfies the equation;

s(t) = (2+ 3√3)/2 = 1 + 3√3/2

What we do here is simply find that value of t that would ensure that;

2sin(t) + 3cos(t) = 1 + 3√3/2

Without any need for rigorous calculations, this value of t can be gotten by inspection.

From our regular trigonometry, we know that the sin of angle 30 is 1/2 and its cos value is √3/2

We can make a substitution for it in this equation.

We obtain the following;

2 sin(30) + 3cos (30) and that is exactly equal to 1 + 3√3/2

Do not forget however that we have a range. And the range in question is between 0 and π/2

Kindly that π/2 in degrees is 90 degrees

So our range of values here is between 0 and 90 degrees.

So to follow the notation in the question, the value within the range that will satisfy the equation is π/6

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3 years ago

What are the zeros of the quadratic function f(x) = 2x2 – 10x – 3?

Natasha2012 [34]

Answer:

$x=\frac{50+\sqrt{31}}{2},\frac{50-\sqrt{31}}{2}$ are zeroes of given quadratic equation.

Step-by-step explanation:

We have been a quadratic equation:

$2x^2-10x-3$

We need to find the zeroes of quadratic equation

We have a formula to find zeroes of a quadratic equation:

$x=\frac{b^2\pm\sqrt{D}}{2a}\text{where}D=\sqrt{b^2-4ac}$

General form of quadratic equation is $ax^2+bx+c$

On comparing general equation with b given equation we get

a=2,b=-10,c=-3

On substituting the values in formula we get

$D=\sqrt{(-10)^2-4(2)(-3)}$

$\Rightarrow D=\sqrt{100+24}=\sqrt{124}$

Now substituting D in $x=\frac{b^2\pm\sqrt{D}}{2a}$ we get

$x=\frac{(-10)^2\pm\sqrt{124}}{2\cdot 2}$

$x=\frac{100\pm\sqrt{124}}{4}$

$x=\frac{100\pm2\sqrt{31}}{4}$

$x=\frac{50\pm\sqrt{31}}{2}$

Therefore, $x=\frac{50+\sqrt{31}}{2},\frac{50-\sqrt{31}}{2}$

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3 years ago

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Lapatulllka [165]

Answer:

does not terminate

Step-by-step explanation:

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3 years ago

Help! I’ll mark you brainly and give extra points!!

irina1246 [14]

Answer:

Slope: 2

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Hope this helps! Can I have Brainliest please?

See attachment below:

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3 years ago

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