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3 years ago

8

F(x)=4x^2-7x+7 Find f(7)

1 answer:

harkovskaia [24]3 years ago

8 0

Answer:

154

Step-by-step explanation:

Because since it says find f(7) you plug the 7 into the x's so it would be...

4(7)^2-7(7)+7= 154

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$6.65

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3 years ago

[14 POINTS AND BRAINLIST ❗️]

julsineya [31]

Answer:

d

Step-by-step explanation:

6 0

3 years ago

How to do foils in algebra

Leno4ka [110]

Foils stands for first, outer, inner, last.

For example:

(3x + 5)(2x + 3)

Multiply the firsts together.

3x * 2x = 5x²

Now multiply the outer together.

3x * 3 = 9x

Now multiply the inners.

5 * 2x = 10x

Now multiply the lasts.

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so the answer would be:

5x² + 9x + 10x + 15
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Hope this helps :)

8 0

2 years ago

The cost price of a car is $60 000. If the marked price is set so as to make a profit of 15%, find the

blsea [12.9K]

Answer:

marked price = $69,000

Step-by-step explanation:

cost price = $60,000

profit

= 15%

= 15% × cost price

= 15/100 × $60,000

= $9000

marked price

= cost price + profit

= $60,000 + $9000

= $69,000

3 0

2 years ago

Consider the differential equation x2y′′ − 9xy′ + 24y = 0; x4, x6, (0, [infinity]). Verify that the given functions form a funda

pantera1 [17]

Answer:

The functions satisfy the differential equation and linearly independent since W(x)≠0

Therefore the general solution is

$y= c_1x^4+c_2x^6$

Step-by-step explanation:

Given equation is

$x^2y'' - 9xy+24y=0$

This Euler Cauchy type differential equation.

So, we can let

$y=x^m$

Differentiate with respect to x

$y'= mx^{m-1}$

Again differentiate with respect to x

$y''= m(m-1)x^{m-2}$

Putting the value of y, y' and y'' in the differential equation

$x^2m(m-1) x^{m-2} - 9 x m x^{m-1}+24x^m=0$

$\Rightarrow m(m-1)x^m-9mx^m+24x^m=0$

$\Rightarrow m^2-m-9m+24=0$

⇒m²-10m +24=0

⇒m²-6m -4m+24=0

⇒m(m-6)-4(m-6)=0

⇒(m-6)(m-4)=0

⇒m = 6,4

Therefore the auxiliary equation has two distinct and unequal root.

The general solution of this equation is

$y_1(x)=x^4$

and

$y_2(x)=x^6$

First we compute the Wronskian

$W(x)= \left|\begin{array}{cc}y_1(x)&y_2(x)\\y'_1(x)&y'_2(x)\end{array}\right|$

$= \left|\begin{array}{cc}x^4&x^6\\4x^3&6x^5\end{array}\right|$

=x⁴×6x⁵- x⁶×4x³

=6x⁹-4x⁹

=2x⁹

≠0

The functions satisfy the differential equation and linearly independent since W(x)≠0

Therefore the general solution is

$y= c_1x^4+c_2x^6$

5 0

3 years ago