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Answer:

133 fishes

Step-by-step explanation:

Units of food A = 400 units

Units of food B = 400 units

Fish Bass required 2 units of A and 4 units of B.

Fish Trout requires 5 units of A and 2 units of B.

i. For food A,

total units of food A required = 2 + 5

                                                 = 7 units

number of bass and trout that would consume food A = 2 x [tex]\frac{400}{7}[/tex]

                                               = 114.3

number of bass and trout that would consume food A = 114

ii. For food B,

total units of food B required = 4 + 2

                                                = 6 units

number of bass and trout that would consume food B = 2 x [tex]\frac{400}{6}[/tex]

                                                 = 133.3

number of bass and trout that would consume food B = 133

Thus, the maximum number of fish that the lake can support is 133.

Answer:

B

Step-by-step explanation:

1) initial circumference

radius * 2 * pi = 24 * pi = 75,398224

2) final circumference

diameter * pi = 48 * pi = 150,796447

150,796447 - 75,398224 = 75.36 m

Answer:

[tex]{ \tt{3 \frac{4}{9} \div (5 \frac{1}{3} - 2 \frac{3}{4}) + 5 \frac{9}{10} }} \\ \\ = { \tt{ \frac{31}{9} \div ( \frac{16}{3} - \frac{11}{4} ) + \frac{59}{10} }} \\ \\ = { \tt{ \frac{31}{9} \div ( \frac{31}{12} ) + \frac{59}{10} }} \\ \\ { \tt{ = \frac{4}{3} + \frac{59}{10} }} \\ \\ { \bf{ = \frac{217}{30} }} \\ \\ { \boxed{ \tt{answer : 7 \frac{7}{30} }}} \\ \\ { \underline{ \blue{ \tt{becker ⚜jnr}}}}[/tex]

Answer:

[tex]7 \frac{7}{30}[/tex]

Step-by-step explanation:

[tex]3 \frac{4}{9} \div ( 5\frac{1}{3} - 2 \frac{3}{4}) + 5 \frac{9}{10}\\\\\frac{31}{9} \div (\frac{16}{3} - \frac{11}{4} ) + \frac{59}{10} \\\\\\Solving \ using \ BODMAS\\\\First \ Solve \ expression \ inside \ Bracket \\\\\frac{31}{9} \div (\frac{(16 \times 4) - ( 11 \times 3)}{12}) + \frac{59}{10} \\\\\frac{31}{9} \div (\frac{64- 33)}{12}) + \frac{59}{10} \\\\\frac{31}{9} \div \frac{31}{12} + \frac{59}{10} \\\\\\ \\\\\\Next \ solve \ Dvision \\\\\frac{\frac{31}{9}}{\frac{31}{12}} + \frac{59}{10}\\\\[/tex]

[tex](\frac{31}{9}} \times {\frac{12}{31}) + \frac{59}{10}[/tex]

[tex]\frac{4}{3} + \frac{59}{10}\\\\ Now \ solve \ final \ expression \\\\\\\frac{(4 \times 10) + ( 59 \times 3)}{30}\\\\\frac{40 + 177}{30}\\\\\frac{217}{30}\\\\7 \frac{7}{30}[/tex]

Step-by-step explanation:

area of a circle is r x r x pi

so one quarter of it us r x r x pi /4

Given:

The function is

[tex]f(x)=(x-5)^2(3-x)^2[/tex]

To find:

The derivative of the given function.

Solution:

Chain rule of differentiation:

[tex][f(g(x))]'=f'(g(x))g'(x)[/tex]

Product rule of differentiation:

[tex][f(x)g(x)]'=f(x)g'(x)+g(x)f'(x)[/tex]

We have,

[tex]f(x)=(x-5)^2(3-x)^2[/tex]

Differentiate with respect to x.

[tex]f'(x)=(x-5)^2\dfrac{d}{dx}(3-x)^2+(3-x)^2\dfrac{d}{dx}(x-5)^2[/tex]

[tex]f'(x)=(x-5)^2[2(3-x)(0-1)]+(3-x)^2[2(x-5)(1-0)][/tex]

[tex]f'(x)=(x^2-10x+25)(-6+2x)+(9-6x+x^2)(2x-10)[/tex]

[tex]f'(x)=(x^2)(-6)+(-10x)(-6)+(25)(-6)+(x^2)(2x)-10x(2x)+25(2x)+(9)(2x)+(-6x)(2x)+x^2(2x)+9(-10)+(-6x)(-10)+x^2(-10)[/tex]

On further simplification, we get

[tex]f'(x)=-6x^2+60x-150+2x^3-20x^2+50x+18x-12x^2+2x^3-90+60x-10x^2[/tex]

[tex]f'(x)=(2x^3+2x^3)+(-6x^2-20x^2-12x^2-10x^2)+(60x+50x+18x+60x)+(-90-150)[/tex]

[tex]f'(x)=4x^3-48x^2+188x-240[/tex]

Therefore, the derivative of the given function is [tex]f'(x)=4x^3-48x^2+188x-240[/tex].

Answer:

0.48780487804

9514 1404 393

Answer:

 1.60

Step-by-step explanation:

The growth factor is 1 more than the growth rate:

  1 + 60% = 1 + 0.60 = 1.60 = growth factor

Answer:

Step-by-step explanation: She has 2 more nickels then dimes not 2 times more therefore answers B and D are incorrect. C is incorrect because it has that there are 2 more dimes than nickels. A is correct because it says that there are c dimes, and then c +2 nickels.

Answer:

x = 14

Step-by-step explanation:

Please note, the word trapezium is a synonym for the word trapezoid.

This problem gives one the area of the trapezoid, a well as one of the measurements of a base and the height of the figure. One is asked to find the length of the other base. This can be done by using the formula to find the area of a trapezoid. This formula is the following,

[tex]A=(h)(\frac{b_1+b_2}{2})[/tex]

Where (A) represents the area of a trapezoid, ([tex]b_1[/tex]) and ([tex]b_2[/tex]) represents the bases and (h) represents the height. Substitute in the given values and solve for the unknown base.

[tex]b_1=7\\h=6\\A=84[/tex]

[tex]A=(h)(\frac{b_1+b_2}{2})\\[/tex]

Substitute,

[tex]84=6(\frac{7+b_2}{2})\\[/tex]

Inverse operations,

[tex]84=6(\frac{7+b_2}{2})[/tex]

[tex]14=\frac{7+b_2}{2}[/tex]

[tex]28=7+b_2[/tex]

[tex]14=b_2[/tex]

Answer:

y = 1.19x

Step-by-step explanation:

y is the dependent variable (total cost)

x is the independent variable (number of pounds)

Answer:

4p² - 9p

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Distributive Property

Algebra I

Step-by-step explanation:

Step 1: Define

Identify

3p(2p - 9) - 2p(-9 + p)

Step 2: Simplify

Answer:

x = 57°

Step-by-step explanation:

Cos(x) = 0.5505

x = Cos⁻¹(0.5505)

x = 56.59

Rounding to the nearest degree;

x = 57°

Hope this helps!

Answer:

= -9pq

Step-by-step explanation:

=8pq + (-17pq)

=8pq-17pq

= -9pq

Let n represent the interger l, the three consective intergers are represented by

[tex]n[/tex]

[tex]n + 2[/tex]

[tex]n + 4[/tex]

The second one represent

[tex](n + 2) {}^{2} + 76 =( n + 4) {}^{2} [/tex]

Simplify both sides

[tex]n {}^{2} + 4n + 4 + 76 = {n}^{2} + 8n + 16[/tex]

[tex] {n}^{2} + 4n + 4 = {n}^{2} + 8n - 60[/tex]

[tex]4n + 4 = 8n - 60[/tex]

[tex]4n + 64= 8n[/tex]

[tex]64= 4n[/tex]

[tex]n = 16[/tex]

The intergers are 16,18,20

Answer:

Step-by-step explanation:

(a - b)(a +b) = a² - b²

1 - Sin² A = Cos² A

[tex]LHS = \frac{1}{1- Sin A} + \frac{1}{1 + Sin A}\\\\= \frac{1*(1 + Sin A)}{(1- Sin A)(1 + Sin A)} + \frac{1*(1- Sin A)}{(1 + Sin A)(1- Sin A)}\\\\= \frac{1 + Sin A+ 1 - Sin A}{1^{2}- Sin^{2} A}\\\\= \frac{2}{1 - Sin^{2} A}\\\\= \frac{2}{Cos^{2} A}\\\\= 2 Sec^{2} A[/tex]

2)  Sec² A - Tan² A = 1

[tex]LHS = \frac{1}{Sec A - Tan A}\\\\=\frac{1*(Sec A + Tan A)}{(Sec A - Tan A)(Sec A + Tan A)}\\\\=\frac{Sec A + Tan A}{Sec^{2} A - Tan^{2} A}\\\\=\frac{Sec A + Tan A }{1}\\\\= Sec A + Tan A = RHS\\\\\\[/tex]

3) LHS  = Cosec² A + Cot² A

             = Cosec² A +  Cosec² A - 1

            = 2Cosec² A - 1   = RHS

[tex]4) LHS = \frac{Sec A}{Cos A}- \frac{Tan A}{Cot A}\\\\ = Sec A*\frac{1}{Cos A}-Tan A*\frac{1}{Cot A}\\\\ = Sec A * Sec A - Tan A * Tan A\\\\= Sec^{2} A - Tan^{2} A \\\\= 1[/tex]

Answer:

The work done is 202.50Nm

Step-by-step explanation:

Given

[tex]F =450N[/tex]

[tex]x_1 = 30cm[/tex]

[tex]x_2 = 60cm[/tex]

Required

The work done

First, we calculate the spring constant (k)

[tex]F = kx_1[/tex]

[tex]450N = k *30cm[/tex]

[tex]k = \frac{450N}{30cm}[/tex]

[tex]k =15N/cm[/tex]

So:

[tex]F = kx_1[/tex]

[tex]F(x) = 15x[/tex]

The work done using Hooke's law is:

[tex]W =\int\limits^a_b {F(x)} \, dx[/tex]

This gives:

[tex]W =\int\limits^{60}_{30} {15x} \, dx[/tex]

Rewrite as:

[tex]W =15\int\limits^{60}_{30} {x} \, dx[/tex]

Integrate

[tex]W =15 \frac{x^2}{2}|\limits^{60}_{30}[/tex]

This gives:

[tex]W =15 *\frac{60^2 - 30^2}{2}[/tex]

[tex]W =15 *\frac{2700}{2}[/tex]

[tex]W =15 *1350[/tex]

[tex]W =20250N-cm[/tex]

Convert to Nm

[tex]W =\frac{20250Nm}{100}[/tex]

[tex]W =202.50Nm[/tex]

Answer:

0.872

Step-by-step explanation:

Given that :

P(cloudy) = P(C) = 0.94

P(cloudy and rainy) = P(C n R) = 0.82

Probability that a given day will be rainy if it is cloudy ; this is a conditional probability problem:

Recall ; P(A|B) = P(AnB) / P(B)

P(R|C) = P(C n R) / P(C) = 0.82 / 0.94 = 0.872

Answer:

-10, -1

Step-by-step explanation:

See Image below:)

Answer:

0.64(x) = 30

Step-by-step explanation:

Hope that's correct.

Answer:

k>1.25

Step-by-step explanation:

The given quadratic equation is :

(k – 1)x² + x + 1 = 0

We need to find all values of k for which the above equation has two real and unequal roots.

For a quadratic equation ax²+bx+c=0, for real and unequal roots,

b²-4ac>0

Here, a = (k-1), b = 1 and c = 1

Put all the values,

1²-4×(k-1)1>0

1-4k+4>0

5-4k>0

k>1.25

S, k can take values more than 1.25. Hence, it can take values 1.5, 1.75.

Answer:

[tex]A_1 = 12[/tex]

[tex]A_n = A_{n-1} + 3[/tex]

Step-by-step explanation:

Given

[tex]A_n =12+(n-1)3[/tex]

Required

Write as recursive

We have:

[tex]A_n =12+(n-1)3[/tex]

Open bracket

[tex]A_n =12+3n-3[/tex]

[tex]A_n =12-3+3n[/tex]

[tex]A_n =9+3n[/tex]

Calculate few terms

[tex]A_1 =9+3*1 = 9 + 3 = 12[/tex]

[tex]A_2 =9+3*2 = 9 + 6 = 15[/tex]

[tex]A_3 =9+3*3 = 9 + 9 = 18[/tex]

The above shows that the rule is to add 3.

So, we have:

[tex]A_1 = 12[/tex]

[tex]A_n = A_{n-1} + 3[/tex]

Answer:

Step-by-step explanation:

Answer:

Arc GH = 78°

Step-by-step explanation:

Inscribed angle = m<GIH = 39°

Measure of arc related to inscribed angle = arc GH = ?

Thus:

m<GIH = ½(arc GH) => Inscribed angles theorem

Substitute

39° = ½(arc GH)

Multiply both sides by 2

2*39° = arc GH

78° = arc GH

Arc GH = 78°

Answer:

A. 11

Step-by-step explanation:

EQ is half of EG

so 22/2 = 11

multiplications

mark me brainliesttt :))

Answer:

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