7. The angle represented by the letter "e" measures 23 degrees. The angle 10 points represented by the letter "s" measures 42 degrees. The angle represent by the letter "pmeasures 67 degrees. Find the angle represented by the letter "d". ONLY A NUMBER WILL BE ACCEPTED. DO NOT PUT ANY OTHER WORDS OR SYMBOLSJUST A NUMBER FOR YOUR ANSWER. FOLLOWING DIRECTIONS IS PART OF THE PROCESS OF PROBLEM SOLVING

[tex]7. The angle represented by the letter e measures 23 degrees. The angle 10 points represented by[/tex]

[tex]t=\frac{a}{\sqrt{1-\frac{y^2}{c^2}}}\\\frac{y^2}{c^2}=1-\frac{a^2}{t^2}\\y^2=\frac{c^2t^2-c^2a^2}{t^2}\\y=\frac{c}{t}\sqrt{t^2-a^2}[/tex]

[tex]t=\frac{a}{\sqrt{1-\frac{v^2}{y^2}}}\\\frac{v^2}{y^2}=1-\frac{a^2}{t^2}\\\\\frac{v^2}{y^2}=\frac{t^2-a^2}{t^2}\\y^2=\frac{v^2t^2}{t^2-a^2}\\y=\frac{vt}{\sqrt{t^2-a^2}}[/tex]

Answer and Explanation:

// code

class Main {

public static void main(String[] args) {

/*

*

*

* your code

*

*/

System.out.println(inDogish("aplderogad"));

System.out.println(inXish("aplderogad", "dog"));

}

// returns true if the word is in dog-ish

// returns false if word is not in dog-ish

public static boolean inDogish(String word) {

// first find d

if (dogishHelper(word, 'd')) {

// first find string after d

String temp = word.substring(word.indexOf("d"));

// find o

if (dogishHelper(temp, 'o')) {

// find string after o

temp = temp.substring(temp.indexOf("o"));

// find g

if (dogishHelper(temp, 'g'))

return true;

}

The output is attached below

}

return false;

}

// necessary to implement inDogish recursively

public static boolean dogishHelper(String word, char letter) {

// end of string

if (word.length() == 0)

return false;

// letter found

if (word.charAt(0) == letter)

return true;

// search in next index

return dogishHelper(word.substring(1), letter);

}

// a generalized version of the inDogish method

public static boolean inXish(String word, String x) {

if (x.length() == 0)

return true;

if (word.length() == 0)

return false;

if (word.charAt(0) == x.charAt(0))

return inXish(word.substring(1), x.substring(1));

return inXish(word.substring(1), x.substring(0));

}

}

PS E:\fixer> java Main true true ne on

PS E:\fixer> java Main true true ne on

[tex]Let’s define a new language called dog-ish. A word is in the lan- guage dog-ish if the word contains[/tex]

1,4&5

Step-by-step explanation:

[tex]d=\frac{C}{\pi}[/tex]

Step-by-step explanation:

Given : C equals pi multiplied by d

We have to solve the equation for d.

Consider the given C equals pi multiplied by d

We first write it mathematically ,

[tex]C=\pi\times d[/tex]

We have to solve for d, that is ,

Divide both side by π, we have,

[tex]\frac{C}{\pi}=d[/tex]

Thus, [tex]d=\frac{C}{\pi}[/tex]

D

Step-by-step explanation:

D is the answer because if you look at the question it is saying "C equals pi times d" so, in this case it would be c=pi/d and none of the answers given is showing that, so therefore D is the answer

d = [tex]\frac{C}{3.14}[/tex]

Step-by-step explanation:

Given

C = 3.14d ( isolate d by dividing both sides by 3.14 )

[tex]\frac{C}{3.14}[/tex] = d

D

Explanation:

This is a letter that is meant to persuade the reader to purchase a certain product.

Answers are mentioned along with explanations/

Explanation:

1. We are given D = 5R

where D = no. of diamonds and R = no. of rubies

a. In words, Number of diamonds = 5 times no. of rubies

In letters D = 5R

b. Given R = 4, so D = 5 x R = 5x 4

D = 20

c. Given R = 10 , so D = 5 x R = 5x 10

D = 50

d. Given D = 15 , D = 5R so , R = D/5 = 15/5 = 3

R = 3

e. R = [tex]\frac{D}{5}[/tex] because D = 5R

2. Given A = B + C

a. B = 5 , C = 2 => A= 5+2 => A= 7

b. B = A - C from above equation and given A = 10 , C = 2 , B= 10 - 2

B = 8

c. two values of B and C for A to be 7

B = 4 C = 3 and vice versa as 4+3 = 7

d. B = A - C by moving C on the RHS thus changing its prefix sign.

3. Given A = 3x and B = 5x+1

a. B goes up faster because it is 5 times x plus 1 where A is just 3 times x

b. if x = 0, A = 3x = 3x0 = > A = 0

c. if x =1 , B = 5x+1 = 5x1 +1 = 5+1 => B = 6

4. Given A = 2x and B = x2 where x2 means x*x so,

a. if x=1 , A = 2*1 = 2 and B = 1*1 = 1

b. if x = 3, A = 3*1 = 3 and B = 3*3 = 9

5. Given D =2R

a. given R = 8 , D = 2*8 => D=16

b. given R =3 then D = 2*3 => D = 6

c. D= 20 given, R = D/2 = 20/2 => R = 10

d. D = 100 so R = D/2 = 100/2 => R =50

e. R = [tex]\frac{D}{2}[/tex] or D/2

6. Given C = π D and π = 3.14

a. if D = 20, C = 3.14*20 => C = 62.8 centimeters

b. if D = 10, C = 3.14*10 => C = 31.4 inches

c. Given C = 18, D = C/31.4 = 18/3.14 => D = 5.732 inches

7. Given S = D/T

a. S = 12/3 = 4 feet/seconds

b. S = 100/5 = 20 miles/seconds

c. For S = 4feet per second , D = 8 ft and T = 2 seconds as S = D/T

we know that

The formula to calculate the circumference of a circle is equal to

[tex]C=\pi d[/tex]

where

d is the diameter of the circle

Solve for d

That means------> isolate the variable d

[tex]C=\pi d[/tex]

Divide by [tex]\pi[/tex] both sides

[tex]d=C/\pi[/tex]

therefore

the answer is

[tex]d=C/\pi[/tex]

34 I am pretty sure

Step-by-step explanation:

34+55 well is 89 so yeah d=34