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2 years ago

10

A fisherman sold some big fish at $4 each and twice as many small fish at

1 answer:

harina [27]2 years ago

5 0

Answer:

The answer is 12 big fish.

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This question has three parts. Answer the parts in order.

nalin [4]

Answer:

The area of the smallest section is $A_{1}=100yd^{2}$

The area of the largest section is $A_{2}=625yd^{2}$

The area of the remaining section is $A_{3}=250yd^{2}$

Step-by-step explanation:

Please see the picture below.

1. First we are going to name the side of the larger square as x.

As the third section shares a side with the larger square and the four sides of a square are equal, we have the following:

- Area of the first section:

$A_{1}=10yd*10yd$

$A_{1}=100yd^{2}$

- Area of the second section:

$A_{2}=x^{2}$ (Eq.1)

- Area of the third section:

$A_{3}=width*length$

$A_{3}=10yd*x$ (Eq.2)

2. The problem says that the total area of the enclosed field is 975 square yards, and looking at the picture below, we have:

$A_{1}+A_{2}+A_{3}=975yd^{2}$

Replacing values:

$100+x^{2}+10x=975$

Solving for x:

$x^{2}+10x-875=0$

$x=\frac{-10+\sqrt{100+(4*875)}}{2}$

$x=\frac{-10+\sqrt{3600}}{2}$

$x=\frac{-10+60}{2}$

$x=25$

3. Replacing the value of x in Eq.1 and Eq.2:

- From Eq.1:

$A_{2}=25^{2}$

$A_{2}=625yd^{2}$

- From Eq.2:

$A_{3}=10*25$

$A_{3}=250yd^{2}$

3 0

2 years ago

78 - 2 4 ÷ ( 14 – 6) × 2 =

Lera25 [3.4K]

Answer:

74

Step-by-step explanation:

you first simplify 2/1 and then 78- (2*2)

6 0

1 year ago

A number times 45 decreased by 12 is no more than 117

Kay [80]

Answer: n*4512 is less than or equal to 117

Step-by-step explanation: n=number N*45=X X=Number more than 117. X decreased by 12 is less than or equal to 117.

5 0

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

5 0

2 years ago

Read 2 more answers

How many real roots does the function

icang [17]

Answer:

Option C is correct i.e. 2.

Step-by-step explanation:

Given the function is f(x) = x² +8x -2.

We can compare it with general quadratic expression i.e. ax² +bx +c.

Then a = 1, b = 8, c = -2.

We can find the number of real root by finding discriminant of the equation ax² +bx +c =0 as follows:-

D = b² -4ac

D = 8² -4*1*-2

D = 64 +8

D = 72.

When D is a positive value, then we have two real roots of the equation.

Hence, option C is correct i.e. 2.

6 0

2 years ago