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3 years ago

F(x)= x^2– 3x + 9 g(x) = 3x^3+ 2x^2– 4x – 9 Find (f - g)(x).

Mathematics

1 answer:

denis-greek [22]3 years ago

5 0

Answer:

$\large \boxed{\sf \ \ -3x^3-x^2+x+18 \ \ }$

Step-by-step explanation:

Hello, please consider the following.

$(f-g)(x)=f(x)-g(x)=x^2-3x+9-(3x^3+2x^2-4x-9)\\\\=x^2-3x+9-3x^3-2x^2+4x+9\\\\=\boxed{-3x^3-x^2+x+18}$

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

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Help me please!

olga nikolaevna [1]

Answer:

Step-by-step explanation:

Well you can get rid of c and d since 9^2 = 81. so 74 is less than 81 so it's definitely less than 9. Now, we know it's between 8 and 9. try multiplying 8.5 x 8.5 = 72.25. So our ranges are

A) 64 - 72.25

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74 falls between 72.25 - 81. So it'll be B

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3 years ago

Which expression is equivalent to ( 5 ^ -2)^5 × 5^4

inna [77]

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3 years ago

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Determine the type and number of solutions of x^2 + 10x − 25 = 0.

Vitek1552 [10]

Answer:

There is exactly 1 solution and it is a double root.

Step-by-step explanation:

To find this, factor the equation and solve.

x^2 + 10x - 25 = 0

(x - 5)(x - 5) = 0

Now that we have this factored, we can set each parenthesis equal to 0 and solve separately.

x - 5 = 0

x = 5

x - 5 = 0

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4 years ago

Can anyone help me find the (?,?) coordinates? The given shape is a parallelogram.

Sonja [21]

The missing coordinates of the parallelogram is (m + h, n).

Solution:

Given shape is a parallelogram.

Construction: Draw a line joining the diagonals.

In parallelogram, diagonals bisect each other.

Solve it using mid-point formula:

$$\text{Midpoint} =\left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right )$

Here $x_1=m, y_1=n, x_2=h, y_2=0$

$$\text{Midpoint} =\left( \frac{m+h}{2}, \frac{n+0}{2}\right )$

$$\text{Midpoint} =\left( \frac{m+h}{2}, \frac{n}{2}\right )$

Using this midpoint find the missing coordinate.

$$\text{Midpoint} =\left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right )$

Let the missing coordinates by x and y.

Here $x_1=0, y_1=0, x_2=x, y_2=y$

$$\text{Midpoint}=\left( \frac{0+x}{2}, \frac{0+y}{2}\right )$

$$\left( \frac{m+h}{2}, \frac{n}{2}\right )=\left( \frac{0+x}{2}, \frac{0+y}{2}\right )$

$$\left( \frac{m+h}{2}, \frac{n}{2}\right )=\left( \frac{x}{2}, \frac{y}{2}\right )$

Now equate the x-coordinates and y-coordinates.

$$ \frac{m+h}{2}=\frac{x}{2}, \ \ \ \frac{n}{2} = \frac{y}{2}$

Multiply by 2 on both sides of the equation, we get

m + h = x, n = y

x = m + h and y = n

Hence the missing coordinates of the parallelogram is (m + h, n).

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3 years ago

Quadrilateral ABCD is reflected over the x-axis to create A′B′C′D′. What are the coordinates of A' ?

Mashcka [7]

When you reflect the figure over the x-axis, you should have a new coordinate of (2,4) for A.

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