radius is 21 5/8 incercepted by 5Pi/6? What is arc lenght?

In an arithmetic sequence, every pair of consecutive terms differs by a fixed number c, so that the n-th term [tex]a_n[/tex] is given recursively by

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

Then for n ≥ 2, we have

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

[tex]a_3=a_2+c = (a_1+c)+c = a_1 + 2c[/tex]

[tex]a_4=a_3+c = (a_1 + 2c) + c = a_1 + 3c[/tex]

and so on, up to

[tex]a_n=a_1+(n-1)c[/tex]

Given that [tex]a_3=126[/tex] and [tex]a_{64}=3725[/tex], we can solve for [tex]a_1[/tex]:

[tex]\begin{cases}a_1+2c=126\\a_1+63c=3725\end{cases}[/tex]

[tex]\implies(a_1+63c)-(a_1+2c)=3725-126[/tex]

[tex]\implies 61c = 3599[/tex]

[tex]\implies c=59[/tex]

[tex]\implies a_1+2\times59=126[/tex]

[tex]\implies a_1+118 = 126[/tex]

[tex]\implies \boxed{a_1=8}[/tex]

Given:

A man owns 3/4 of the share of a business and sells 1/3 of his  shares for Bir 10,000.

To find:

The value of the business in Bir.

Solution:

Let x be the value of the business.

It is given that a man owns 3/4 of the share of a business and sells 1/3 of his  shares for Bir 10,000.

[tex]x\times \dfrac{3}{4}\times \dfrac{1}{3}=10000[/tex]

[tex]x\times \dfrac{1}{4}=10000[/tex]

Multiply both sides by 4.

[tex]x=4\times 10000[/tex]

[tex]x=40000[/tex]

Therefore, the value of the business is 40,000 Bir.

Answer:

C. 11x² - 3x

Step-by-step explanation:

(12x² - 5x) - (x² - 2x)

12x² - 5x - x² + 2x

12x - x² - 5x + 2x

11x² - 3x

Answer:

3/11

Step-by-step explanation:

Answer:4 ma boi

Step-by-step explanation:

Answer:

14

Step-by-step explanation:

because I am god at meth and very smart

Answer:

ik only 1 of them and i dont even know if this is correct...

a) A

Answer:

y = 2x + 10

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Here m = 2 , then

y = 2x + c ← is the partial equation

To find c substitute (- 4, 2 ) into the partial equation

2 = - 8 + c ⇒ c = 2 + 8 = 10

y = 2x + 10 ← equation of line

Answer:

72 in. sq.

Step-by-step explanation:

6 x 6 = 36

6 x 6 = 36 / 2 = 18

6 x 6 = 36 / 2 = 18

18 + 18 + 36 = 72.

Hope this helps!

If there is something wrong just let me know.

Step-by-step explanation:

here's the answer to your question

Answer: Distance = 11

Step-by-step explanation:

Concept:

Here, we need to know the idea of the distance formula.

The distance formula is the formula, which is used to find the distance between any two points.

If you are still confused, please refer to the attachment below for a clear version of the formula.

Solve:

Find the distance between A and B, where:

[tex]Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]

[tex]Distance=\sqrt{(4+7)^2+(-4+4)^2}[/tex]

[tex]Distance=\sqrt{(11)^2+(0)^2}[/tex]

[tex]Distance=\sqrt{121+0}[/tex]

[tex]Distance=\sqrt{121}[/tex]

[tex]Distance=11[/tex]

Hope this helps!! :)

Please let me know if you have any questions

Answer:

A) quadratic

Step-by-step explanation:

ax2 + bx+c=0

Since the highest power of the equation is 2

A) quadratic -2

B) quartic- 4

C) linear- 1

D) cubic-3

Answer:

201.088 cubic units

Step-by-step explanation:

Given data

height= 12 units

radius= 4 units

The expression for the volume of a cone is given as

Volume = 1/3*πr^2h

Substitute the given data to find the volume of the cone

Volunme= 1/3* 3.142* 4^2* 12

Volume = 1/3* 3.142* 16*12

Volume= 1/3* 603.264

Volume= 201.088 cubic units

Answer:

a = -6

b = 1

Step-by-step explanation:

The gradient of the tangent to the curve y = ax + bx^3, will be:

dy/dx = a + 3bx²

at (2, -4)

dy/dx = a+3b(2)²

dy/dx = a+12b

Since the gradient at the point is 6, then;

a+12b = 6 ....1

Substitute x = 2 and  y = -4 into the original expression

-4 = 2a + 8b

a + 4b = -2 ...2

a+12b = 6 ....1

Subtract

4b - 12b = -2-6

-8b = -8

b = -8/-8

b = 1

Substitute b = 1 into equation 1

Recall from 1 that a+12b = 6

a+12(1) = 6

a = 6 - 12

a = -6

Hence a = -6, b = 1

Answer:

please someone help me with my latest question

Answer:

41 units

Step-by-step explanation:

Its the reflection the length are same

Answer:

The solutions are w=0 ,7

Step-by-step explanation:

3w^2 – 21w = 0

Factor out 3w

3w(w-7) =0

Using the zero product property

3w=0  w-7=0

w =0    w=7

The solutions are w=0 ,7

Answer:

the number before the first variable (first term)

Step-by-step explanation:

this appears to be an incomplete question. The numerical coefficient of a term is the number before the variable.

the constant is the number without a variable.

Step-by-step explanation:

You didn't put the graph, but you can compare between your graphs and the picture.

Brainliest please

The graph that represents this inequality y ≥ 4x − 3 is attached below.

If the equation or inequality contains variable terms, then there might be some values of those variables for which that equation or inequality might be true. Such values are called solution to that equation or inequality. Set of such values is called solution set to the considered equation or inequality.

We are given that the inequality is;

y ≥ 4x − 3

The slope of the inequality is 4.

The equation of the red line is y = 4x − 3  

The shading is above the line and the line is solid, that means  y is greater than or equal 4x − 3

The graph of this inequality y ≥ 4x − 3 is attached below.

Learn more about inequalities here:

https://brainly.com/question/27425770

#SPJ2

Answer:

Complementary angle

<y+61° =90°

<y = 90°- 61°

<y =29°

I Hope this helps.

Answer:

Step-by-step explanation:

5t² + 4t = 5t² + 4 if and only if t=1.

A zero coefficient makes the value of the term equal to zero.

Answer: 5t to the second power and 5t to the second power. I don't know the answer to b.

Step-by-step explanation:

Answer:

x = 2 - sqrt(19/6)

x = 2 + sqrt(19/6)

Step-by-step explanation:

Answer:

add 4

subtract 24 from 5

2

Step-by-step explanation:

We know

[tex]\boxed{\sf P(x,y)=\left(\dfrac{mx_2+nx_1}{m+n},\dfrac{my_2+ny_1}{m+n}\right)}[/tex]

[tex]\\ \sf\longmapsto p(x,y)=\left(\dfrac{3(7)+5(-1)}{3+5},\dfrac{3(7)+5(1)}{3+5}\right)[/tex]

[tex]\\ \sf\longmapsto p(x,y)=\left(\dfrac{21-5}{8},\dfrac{21+5}{8}\right)[/tex]

[tex]\\ \sf\longmapsto p(x,y)=\left(\dfrac{16}{8},\dfrac{26}{8}\right)[/tex]

[tex]\\ \sf\longmapsto p(x,y)=\left(2,\dfrac{13}{4}\right)[/tex]

m:n=3:5

We know

[tex]\boxed{\sf M(x,y)=\left(\dfrac{mx_2+nx_1}{m+n},\dfrac{my_2+ny_1}{m+n}\right)}[/tex]

[tex]\\ \sf\longmapsto M(x,y)=\left(\dfrac{3(7)+5(-1)}{3+5},\dfrac{3(7)+5(1)}{3+5}\right)[/tex]

[tex]\\ \sf\longmapsto M(x,y)=\left(\dfrac{21-5}{8},\dfrac{21+5}{8}\right)[/tex]

[tex]\\ \sf\longmapsto M(x,y)=\left(\dfrac{16}{8},\dfrac{26}{8}\right)[/tex]

[tex]\\ \sf\longmapsto M(x,y)=\left(2,\dfrac{13}{4}\right)[/tex]

Answer:

There is significant evidence to conclude that carina e in New section is greater than 42.3

Step-by-step explanation:

Given :

Sample variance, s² = 48.3

Population variance, σ² = 42.3

Sample size, n = 28

α = 0.05

The hypothesis :

H0 : σ² = 42.3

H0 : σ² > 42.3

The test statistic, χ² : (n-1)*s²/σ²

χ² = [(28 - 1) * 48.3] / 42.3

χ² = (27 * 48.3) / 42.3

χ² = 1304.1 / 42.3

χ² = 30.829787

χ² = 30.830

The Critical value at α = 0.05 ; df = (28-1) = 27

Critical value(0.05, 27) = 27.587

Reject H0 if χ² statistic > Critical value

Since, 30.830 > 27.587 ; Reject H0 ; and conclude that variance in the new section is greater than 42.3

Answer:The value of x is 4.

Step-by-step explanation:

Given : Expression  .

To find : Solve for x ?

Solution :

Applying exponential property,  

Comparing the base,

Step-by-step explanation:

Answer:

VERY NICE RACK U HAVE MAM

Step-by-step explanation:

Answer:

Its a

Step-by-step explanation:

found on another thing and im taking test

Answer:

a) Hence the probability that the average test score in the class of size 25 exceeds 80.

P ( X > 74) = 0.9838

P ( Z > 2.14) = 0.0162

b) Hence the probability that the average test score for the class of size 64

P ( X > 74) = 0.9838

P ( Z > 2.14) = 0.0003

c) Probability of the difference exceeding 2.2 = 0.9936

P (Z < 2.49) = 0.0064

Step-by-step explanation:

Let's assume a normal distribution.

Now,

a) For a class of 25

[tex]P ( X > 74) = P (Z > 80 - 74/ 14\sqrt(25) )\\\\= P (Z < 6 / 2.8)\\\\= P ( Z < 2.14)\\= 0.9838\\[/tex]

[tex]P ( Z > 2.14) = 1- 0.9838\\= 0.0162[/tex]

b)

Similarly:

For the class of 64

[tex]P ( X > 74) = P (Z > 80 - 74/ 14\sqrt(64) )\\\\= P (Z < 6 / 1.75)= P ( Z < 3.428)\\= 0.9838\\[/tex]

[tex]P ( Z > 2.14) = 1- 0.9997\\= 0.0003[/tex]

c) Probability of the difference exceeding 2.2

[tex]= P (Z > 2.2/\sqrt{14 * {(1/25) + (1/64}\\[/tex]

[tex]= P (Z > 2.49)\\= 0.9936[/tex]

P (Z < 2.49)

= 1 - 0.9936

= 0.0064

Answer:

sorry I don't know the answer

Answer:

For the equation y=log(x-k), the domain depends on the value of K. Sliding K moves the left bound of the domain interval. The range and the right end behavior stay the same. For the equation y=log x+k, the domain is fixed, starting at an x-value of 0. The vertical asymptote is also fixed. The range of the equation depends on K.

Step-by-step explanation:

Answer:

The numerical limits for a C grade are 60.6 and 69.1.

Step-by-step explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Scores on the test are normally distributed with a mean of 67 and a standard deviation of 7.3.

This means that [tex]\mu = 67, \sigma = 7.3[/tex]

Find the numerical limits for a C grade.

Below the 100 - 38 = 62th percentile and above the 18th percentile.

18th percentile:

X when Z has a p-value of 0.18, so X when Z = -0.915.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]-0.915 = \frac{X - 67}{7.3}[/tex]

[tex]X - 67 = -0.915*7[/tex]

[tex]X = 60.6[/tex]

62th percentile:

X when Z has a p-value of 0.62, so X when Z = 0.305.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]0.305 = \frac{X - 67}{7.3}[/tex]

[tex]X - 67 = 0.305*7[/tex]

[tex]X = 69.1[/tex]

The numerical limits for a C grade are 60.6 and 69.1.

Answer:

D. x = 4,  y = 4, z = 2

Step-by-step explanation

Plugged in given answers as trying to substitute is impossible, already tried all combinations

Solution :

Given :

2y''' + 15y'' + 24y' + 11y= 0

Let x = independent variable

[tex](a_0D^n + a_1D^{n-1}+a_2D^{n-2} + ....+ a_n) y) = Q(x)[/tex]  is a differential equation.

If [tex]Q(x) \neq 0[/tex]

It is non homogeneous then,

The general solution  = complementary solution + particular integral

If Q(x) = 0

It is called the homogeneous then the general solution =  complementary solution.

2y''' + 15y'' + 24y' + 11y= 0

[tex]$(2D^3+15D^2+24D+11)y=0$[/tex]

Auxiliary equation,

[tex]$2m^3+15m^2+24m +11 = 0$[/tex]

-1  | 2    15    24     11

    | 0   -2    - 13    -11  

      2    13    11       0

∴ [tex]2m^2+13m+11=0[/tex]

The roots are

[tex]$=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$[/tex]

[tex]$=\frac{-13\pm \sqrt{13^2-4(11)(2)}}{2(2)}$[/tex]

[tex]$=\frac{-13\pm9}{4}$[/tex]

[tex]$=-5.5, -1$[/tex]

So, [tex]m_1, m_2, m_3 = -1, -1, -5.5[/tex]

Then the general solution is :

[tex]$= (c_1+c_2 x)e^{-x} + c_3 \ e^{-5.5x}$[/tex]

Answer:

The number of ways of selecting the team is 26,400 ways.

Step-by-step explanation:

Given;

total number boys in the gym, b = 10 boys

total number of girls in the gym, g = 12 girls

number of team to be selected, n = 6

If there must equal number of boys and girls in the team, then the team must consist of 3 boys and 3 girls.

Number of ways of choosing 3 boys from the total of 10 = [tex]10_C_3[/tex]

Number of ways of choosing 3 girls from a total of 12 = [tex]12_C_3[/tex]

The number of ways of combining the two possibilities;

[tex]n = 10_C_3 \times 12_C_3\\\\n = \frac{10!}{7!3!} \ \times \ \frac{12!}{9!3!} \\\\n = \frac{10\times 9 \times 8}{3\times 2} \ \times \ \frac{12\times 11 \times 10}{3\times 2} \\\\n = 120 \times 220\\\\n = 26,400 \ ways[/tex]

Therefore, the number of ways of selecting the team is 26,400 ways.