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3 years ago

1 11 f(x) = 2x² – 7x - 1 g(x) = 4x + 5 Find: go f(8)

Mathematics

1 answer:

Snezhnost [94]3 years ago

6 0

Answer: 71

Step-by-step explanation:

F(8)= 2(8)^2-7(8)-1

=71

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The price of a company's share dropped by 3.50% by the end of the first year, down to $44.25. During the second year the price o

omeli [17]

Answer:

a. $45.86

b. $43.87

c. Percentage change = -4.40%)

Step-by-step explanation:

a. Let the price of share at the beginning of first year be represented with X

Now, Price of company's share dropped by 3.50% at the end of first year.

So, Price of share at the end of first year = x - 3.5% of x

= x - 0.035x

= 1x - 0.035x

= 0.965x

But it is equal to $44.25

=> 0.965x = 44.25

x = 44.25 / 0.965

x = 45.8549223

x = $45.86

b. During second year price of share decreased by $1.99. Therefore, Price of share at the end of second year = $45.86 - $1.99 = $43.87

c. Percentage change in the price of shares over two years = {(45.86 - 43.87)/45.86} *100

= (1.99/45.86)*100

= 0.04339294 * 100

= 4.40%

Now, as price of shares has dropped, the percentage change will be negative. (Δ% = -4.40%)

5 0

2 years ago

. what is the area of the pentagon shown below?

Ronch [10]

Answer:

About 80.5 feet.

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

Solve 7 ÷ one half = ___. Question 1 options: 14 15 24 40

Norma-Jean [14]

Answer:

O 14

Step-by-step explanation:

hope it is helpful :)

7 0

2 years ago

So far this month John calculated that he will earn $40 in comission for his $500 in product sales. If John would like his comis

Helen [10]

John made $40 commissions for $500 sales or 40/500. He wants $320 for an unknown amount of sales or 320/x.

40/500=320/x Cross multiply.

40x=160,000

Divide both sides by 40.

X=4,000

He needs his sales to be $4,000 to get $320 in commission.

4 0

3 years ago

For each of the following vector fields

olga nikolaevna [1]

(A)

$\dfrac{\partial f}{\partial x}=-16x+2y$

$\implies f(x,y)=-8x^2+2xy+g(y)$

$\implies\dfrac{\partial f}{\partial y}=2x+\dfrac{\mathrm dg}{\mathrm dy}=2x+10y$

$\implies\dfrac{\mathrm dg}{\mathrm dy}=10y$

$\implies g(y)=5y^2+C$

$\implies f(x,y)=\boxed{-8x^2+2xy+5y^2+C}$

(B)

$\dfrac{\partial f}{\partial x}=-8y$

$\implies f(x,y)=-8xy+g(y)$

$\implies\dfrac{\partial f}{\partial y}=-8x+\dfrac{\mathrm dg}{\mathrm dy}=-7x$

$\implies \dfrac{\mathrm dg}{\mathrm dy}=x$

But we assume $g(y)$ is a function of $y$ alone, so there is not potential function here.

(C)

$\dfrac{\partial f}{\partial x}=-8\sin y$

$\implies f(x,y)=-8x\sin y+g(x,y)$

$\implies\dfrac{\partial f}{\partial y}=-8x\cos y+\dfrac{\mathrm dg}{\mathrm dy}=4y-8x\cos y$

$\implies\dfrac{\mathrm dg}{\mathrm dy}=4y$

$\implies g(y)=2y^2+C$

$\implies f(x,y)=\boxed{-8x\sin y+2y^2+C}$

For (A) and (C), we have $f(0,0)=0$ , which makes $C=0$ for both.

4 0

3 years ago