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3 years ago

ACB=67 degrees find BCD

Mathematics

1 answer:

Sunny_sXe [5.5K]3 years ago

4 0

Answer:

23°

Step-by-step explanation:

Since ACB is a right angle, the two angles ACB and BCD add up to 90°

just subtract 67 from 90 and bam 23

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Find the missing factor BBB that makes the equality true.

loris [4]

Answer:21y^4= B*7y^3

B=(21y^4)/(7y^3)

B=3y

i think this is correct

4 0

3 years ago

After completing the square, what are the solutions to the quadratic equation

Fiesta28 [93]

Answer:

$x_1= \frac{\sqrt{{29}}}{2} +\frac{7}{2}\\\\x_2= -\frac{\sqrt{{29}}}{2} +\frac{7}{2}$

Step-by-step explanation:

1. Subtract 9 from both sides of the equation:

$x^2 - 7x+9-9 = 4-9\\\\x^2 - 7x= -5$

2. Notice that the coefficient of "x" is 7. Then:

$(\frac{7}{2})^2=\frac{49}{4}$

3. Add $\frac{49}{4}$ to both sides of the equation:

$x^2 - 7x+\frac{49}{4}= \frac{49}{4}-5\\\\x^2 - 7x+\frac{49}{4}= \frac{29}{4}$

4. Completing the square, you get:

$(x -\frac{7}{2})^2= \frac{29}{4}$

5. To find the solutions, you must square-root both sides of the equation, simplify, and solve for "x". Then, you get:

$\sqrt{(x -\frac{7}{2})^2}= \±\sqrt{\frac{29}{4}}\\\\x -\frac{7}{2}=\±\frac{\sqrt{{29}}}{2} \\\\x= \±\frac{\sqrt{{29}}}{2} +\frac{7}{2}\\\\\\x_1= \frac{\sqrt{{29}}}{2} +\frac{7}{2}\\\\x_2= -\frac{\sqrt{{29}}}{2} +\frac{7}{2}$

4 0

3 years ago

Read 2 more answers

May someone help me out with this please?

bixtya [17]

I think it Would it be 260?

5 0

4 years ago

a cookie jar contains four oatmeal raisin, one sugar, nine chocolate chip, and six peanut butter cookies. if one is chosen at ra

VikaD [51]

Step-by-step explanation:

There are a total of 4 + 1 + 9 + 6 = 20 cookies. So the probabilities of each type for a random cookie are:

P(oatmeal raisin) = 4/20 = 1/5

P(sugar) = 1/20

P(chocolate chip) = 9/20

P(peanut butter) = 6/20 = 3/10

8 0

3 years ago

Which ordered pair is the solution to the system of equations?

d1i1m1o1n [39]

$\begin{cases}-3x+4y=-20\\y=x-4\end{cases}\\\\\\-3x+4(x-4)=-20\\\\-3x+4x-16=-20\\\\x-16=-20\quad|+16\\\\x-16+16=-20+16\\\\\boxed{x=-4}\\\\\\y=x-4\\\\y=-4-4\\\\\boxed{y=-8}$

Answer C.

4 0

3 years ago