Ask question

Ask question

All categories

3 years ago

12

If x = 2 is a zero of the polynomial

" title="2x {}^{2} + 3x - p" alt="2x {}^{2} + 3x - p" align="absmiddle" class="latex-formula">, then find the value if p.

1 answer:

Vladimir79 [104]3 years ago

4 0

Given : $f(x) = 2x {}^{3} + 3x - p$

As x = 2 is a zero of the polynomial
So, $f(2) = 0$
$2.2 {}^{3} + 3.2 - p = 0 = > 16 + 6 - p = 0$
$= > 22 - p = 0 = > p = 22$

You might be interested in

Mrs. Duncan pours

trapecia [35]

Answer:

divide 8.4 with 3

the answer is 2.8

4 0

3 years ago

Read 2 more answers

If I use the Quotient of Powers Property to simplify this expression, I should ( add, subtract, multiply, divide) the exponents.

Makovka662 [10]

7^8 / 7^6 = 7^2 = 49
The formula is [a^m / a^n = a^m-n]
I hope this helped :)

7 0

3 years ago

Convert 9 1/2 in to a fractional notation.

Evgen [1.6K]

2 x 9 + 1 = 19

19/2

Thus, the answer is 19/2

6 0

3 years ago

Read 2 more answers

This table shows a linear relationship.

svp [43]

The answer would be y=42x

5 0

2 years ago

An aircraft manufacturer wants to determine the best selling price for a new airplane. The company estimates that the initial co

Blizzard [7]

Answer:

$(a)$ $C(x)=500+20x-5x^{\frac{3}{4}}+0.01x^2$

$p(x)=320-7.7p$

$R(x)=(320-7.7p)p=320p-7.7p^2$

$(b)$ $x=82 \text{planes}$

$(c)$ $p=\$30.91 M\;\; \text{per plane}$

$(d)$ maximum profit $=\$ 15.90M$

Step-by-step explanation:

Given that,

The company estimates that the initial cost of designing the aeroplane and setting up the factories in which to build it will be 500 million dollars.

The additional cost of manufacturing each plane can be modelled by the function.

$m(x)=20x-5x^{\frac{3}{4}}+0.01x^2$

$(a)$ Find the cost, demand (or price), and revenue functions.

$C(x)=500+20x-5x^{\frac{3}{4}}+0.01x^2$

$p(x)=320-7.7p$

$R(x)=(320-7.7p)p=320p-7.7p^2$

$(b)$ Find the production level that maximizes profit.

$f=R(x)-C(x)$

$\Rightarrow f=320p-7.7p^2-(500+20x-5x^{\frac{3}{4}}+0.01x^2)$

$\Rightarrow df=320dp-15.4pdp-20dx+5(\frac{3}{4} )x^{\frac{-1}{4} }dx-0.02xdx$

$x=320-7.7p$

$p=\frac{320-x}{7.7}$

$\frac{dp}{dx} = \frac{-1}{7.7}$

$\frac{df}{dx}=\frac{320}{-7.7} -\frac{15.4(320-x) }{7.7(\frac{-1}{7.7} )}-20+5\frac{3}{4} x^{\frac{-1}{4}} -0.02x=0$

$\Rightarrow -41.5584+83.1169-0.2597x-20+3.75x^{\frac{-1}{4} }-0.02x=0$

$\Rightarrow 21.5585+3.75x^{\frac{-1}{4} }-0.279x=0$

$\Rightarrow x=82 \text{planes}$

$(c)$ Find the associated selling price of the aircraft that maximizes profit.

$p=\frac{320-82}{7.7}$

$\Rightarrow p=\$30.91 M\;\; \text{per plane}$

$(d)$ Find the maximum profit.

Manufacturing cost of one plane is:

$m(1)=20-5+0.01$

$=\$15.01 M$

maximum profit $=\$(30.91-15.01)M$

$=\$15.90M$

3 0

3 years ago