The times taken by 18 people to complete a puzzle are shown

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:

y = 1.19x

Step-by-step explanation:

y is the dependent variable (total cost)

x is the independent variable (number of pounds)

Answer:

The value of h(g(3)) is 17.

Step-by-step explanation:

We are given these following functions:

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

[tex]h(x) = 7x - 11[/tex]

h(g(3)) ?

[tex]h(g(x)) = h(x^2-5) = 7(x^2-5) - 11 = 7x^2 - 35 - 11 = 7x^2 - 46[/tex]

At x = 3

[tex]h(g(3)) = 7(3)^2 - 46 = 63 - 46 = 17[/tex].

The value of h(g(3)) is 17.

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.

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:

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:

0 + 0.5, 0 + 0.6 + 0.05, 2 + 0.3 + 0.05, 1 + 0.06

Step-by-step explanation:

expanded form is just taking the number apart from ones to tenths to hundreds and so on, or from ones to tens to hundreds and then so on. then you just split it and put the numbers in added form.

Answer:

A. 11

Step-by-step explanation:

EQ is half of EG

so 22/2 = 11

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:

= -9pq

Step-by-step explanation:

=8pq + (-17pq)

=8pq-17pq

= -9pq

Answer:

It would be the letter B :)

Answer:

Step-by-step explanation:

Answer:

I think B

Step-by-step explanation:

121.346° is more close to 121.3°, than 121.4°

if i'm wrong, the i'm sorry

Answer:

0.64(x) = 30

Step-by-step explanation:

Hope that's correct.

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:

[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]

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:

68

Step-by-step explanation:

∠ACB = ∠DCE

          = 180-92/2

          =44

∠CDE = 180 - 44 / 2

         68

Answer:

x = 3 + 2y

Step-by-step explanation:

x - 2y = 3 ( Isolate x on the left side by adding 2y to both sides )

x = 3 + 2y

Step-by-step explanation:

Step 1: Add -y^3 to both sides.

y3+x2+−y3=1+−y3

x2=−y3+1

Step 2: Take square root.

x=√−y3+1 or x=−√−y3+1

Answer:

x=√−y3+1  or  x=−√−y3+1

(I think this is right) (tell me if im right plz and thx)

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

[tex]\sf\purple{A.\:Between \:12\:and\:13.}[/tex] ✅

[tex]\large\mathfrak{{\pmb{\underline{\red{Step-by-step\:explanation}}{\orange{:}}}}}[/tex]

[tex] \sqrt{156} \\ = 12.4899 \\ = 12.49[/tex]

Therefore, [tex] \sqrt{156} [/tex] will fall in between 12 and 13.

[tex]\large\mathfrak{{\pmb{\underline{\orange{Happy\:learning }}{\orange{!}}}}}[/tex]

Answer:

-10, -1

Step-by-step explanation:

See Image below:)

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.

220 cubic units.

Answer:

Solution given:

for small cylinder

r=1

and for large cylinder

R=5+1=6

height for both [h]=2

Now

Volume of solid=πR²h-πr²h=πh(R²-r²)

=3.14*2(6²-1²)=219.8 =220 units ³.

Small cylinder is r=1

Large cylinder is R= 5+1 =6

Height (h) =2

Volume of solid,

→ πR²h-πr²h

→ πh(R²-r²)

→ 3.14 × 2(6²-1²)

→ 219.8

→ 220 cubic units

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]