Someone pls help…. We need to find m angle f

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:

B.10

Step-by-step explanation:

15=1/2b(3)

30=3b

10= base

9514 1404 393

Answer:

Step-by-step explanation:

1. The finance charge is found from the simple interest formula;

  I = Prt

where P is the principal amount, r is the annual rate, and t is the number of years.

24 months is 2 years, so the interest charged is ...

  I = $1900×0.08×2 = $304

The finance charge is $304.

__

2. The monthly payment will be the total amount due, divided by the number of months.

  payment = ($1900 +304)/24 = $2204/24 ≈ $91.83

The monthly payment is $91.83.

I hope it will help you.

Answer:

x = 9

Step-by-step explanation:

angle <1 and angle <2 is complementary and their sum is 90 degrees

3x + 5x + 18 = 90 add like terms

8x + 18 = 90 subtract 18 from both sides

8x = 72 divide both sides by 8

x = 9 to find the measure of angle <1 replace x with the value we found

[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:

[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:

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:

x = 6 , y = 7

Step-by-step explanation:

solving  by substitution method

x + y = 13

x = 13 - y                          equation (i)

2x - y = 5

substitute the value of x

2(13 - y) - y = 5

26 - 2y - y = 5

26 - 3y = 5

26 - 5 = 3y

21/3 = y

7 = y

substitute the value of y in equation (i)

x = 13 - y

x = 13 - 7

x = 6

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:

Step-by-step explanation:

Soooo, by looking at the picture we can tell that it is a right trapezoid

The formula for the area is

(b1+b2)/2*h

(Base 1+ Base 2)/ 2 * height

Now we can plug in the numbers:

(5+15)/2*6

20/2*6

10*6

=60

1 gallon of paint per sq. foot

So Isha needs 60 gallons of paint

Answer:

09

Step-by-step explanation:

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:

-10, -1

Step-by-step explanation:

See Image below:)

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:

= -9pq

Step-by-step explanation:

=8pq + (-17pq)

=8pq-17pq

= -9pq

Answer:

-25

Step-by-step explanation:

-24×(-25)=600

Hope this helps! :)

Answer: It's -25

edg 2023

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:

Well, 10% of 6640 is $664, and $332 is half of that, so 5%

Missing: ​$9850 ‎​$591.

Step-by-step explanation:

Answer:

espero ayudarte ..............

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:

[tex]a_1 = 2[/tex]

[tex]a_2 = 3[/tex]

[tex]a_3 = 5[/tex]

[tex]a_4 = 8[/tex]

[tex]a_5 = 13[/tex]

Step-by-step explanation:

Given

[tex]a_n = a_{n-1} + a_{n-2}[/tex]

[tex]a_1 = 2[/tex]

[tex]a_2 = 3[/tex]

Solving (a): The first term

This has already been given as:

[tex]a_1 = 2[/tex]

Solving (b): The second term

This has already been given as:

[tex]a_2 = 3[/tex]

Solving (c): The third term

This is calculated as:

[tex]a_n = a_{n-1} + a_{n-2}[/tex]

[tex]a_3 = a_{3-1} +a_{3-2}[/tex]

[tex]a_3 = a_2 +a_1[/tex]

[tex]a_3 = 3 +2[/tex]

[tex]a_3 = 5[/tex]

Solving (d): The fourth term

This is calculated as:

[tex]a_n = a_{n-1} + a_{n-2}[/tex]

[tex]a_4 = a_{4-1} +a_{4-2}[/tex]

[tex]a_4 = a_3 +a_2[/tex]

[tex]a_4 = 5+3[/tex]

[tex]a_4 = 8[/tex]

Solving (e): The fifth term

This is calculated as:

[tex]a_n = a_{n-1} + a_{n-2}[/tex]

[tex]a_5 = a_{5-1} +a_{5-2}[/tex]

[tex]a_5 = a_4 +a_3[/tex]

[tex]a_5 = 8+5[/tex]

[tex]a_5 = 13[/tex]

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:

0.64(x) = 30

Step-by-step explanation:

Hope that's correct.

Answer:

y = 1.19x

Step-by-step explanation:

y is the dependent variable (total cost)

x is the independent variable (number of pounds)

Answer:

Step-by-step explanation:

Answer:

(6290.678 ; 7790.742)

Step-by-step explanation:

Given the data :

5640, 5090, 6590, 6380, 7165, 8440, 9980

The sample mean, xbar = Σx / n = 49285 / 7 = 7040.71

The 90% confidence interval :

Xbar ± Margin of error

Margin of Error = Zcritical * σ/√n

Since the σ is known, we use the z- distribution

Zcritical at 90% confidence = 1.64

Hence,

Margin of Error = 1.64 * 1210/√7

Margin of Error = 750.032

90% confidence interval is :

7040.71 ± 750.032

Lower boundary = 7040.71 - 750.032 = 6290.678

Upper boundary = 7040.71 + 750.032 = 7790.742

(6290.678 ; 7790.742)