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kvasek [131]
3 years ago
9

I need help on this math problem.

Mathematics
1 answer:
OLEGan [10]3 years ago
6 0

B: (0, a)

C should have the same y value as B

D: (a,0)

C should have the same x value as D


So point C is (a,a)

Since point A is on the origin, its point is (0,0)


You use the slope formula and plug in point A and C:

m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}

m=\frac{a-0}{a-0}

m=\frac{a}{a}

m = 1


So the value that belongs in the green box is 1


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EPCIUCNO
cluponka [151]

Answer:

1 ⇒ D

2 ⇒ H

3 ⇒ B

4 ⇒ G

5 ⇒ A

6 ⇒ F

7 ⇒ E

8 ⇒ C

Step-by-step explanation:

* Lets explain how to solve the problem

- The left column has expressions and the right column has the

  equivalent expressions

- We must find the equivalent letter for each number

1.

# 7(12)

∵ 12 can be a sum of two numbers

∴ The equivalent expression to 7(12) is 7(8 + 4)

∴ 1 ⇒ D

2.

# 3(15)

∵ 3 can be a sum of two numbers

∴ The equivalent expression to 3(15) is (2 + 1)(15)

∴ 2 ⇒ H

3.

# 3a + 9

∵ 3 and 9 have a common factor 3

∵ 3a + 9 ⇒ divide them by 3

∵ 3a ÷ 3 = a and 9 ÷ 3 = 3

∴ 3a + 9 = 3(a + 3)

∴ The equivalent expression to 3a + 9 is 3(a + 3)

∴ 3 ⇒ B

4.

# 9a + 3

∵ 9 and 3 have a common factor 3

∵ 9a + 3 ⇒ divide them by 3

∵ 9a ÷ 3 = 3a and 3 ÷ 3 = 1

∴ 9a + 3 = 3(3a + 1)

∴ The equivalent expression to 9a + 3 is 3(3a + 1)

∴ 4 ⇒ G

5.

# 5 + 15a

∵ 5 and 15 have a common factor 5

∵ 5 + 15a ⇒ divide them by 5

∵ 5 ÷ 5 = 1 and 15a ÷ 5 = 3a

∴ 5 + 15a = 5(1 + 3a)

∴ The equivalent expression to 5 + 15a is 5(1 + 3a)

∴ 5 ⇒ A

6.

# 10 + 5a

∵ 10 and 5 have a common factor 5

∵ 10 + 5a ⇒ divide them by 5

∵ 10 ÷ 5 = 2 and 5a ÷ 5 = a

∴ 10 + 5a = 5(2 + aa)

∴ The equivalent expression to 10 + 5a is 5(2 + a)

∴ 6 ⇒ F

7.

# 3x + 6y + 9z

∵ The coefficient of x , y , z have a common factor 3

∵ 3x ÷ 3 = x

∵ 6y ÷ 3 = 2y

∵ 9z ÷ 3 = 3z

∴ 3x + 6y + 9z = 3(x + 2y + 3z)

∴ The equivalent expression to 3x + 6y + 9z is 3(x + 2y + 3z)

∴ 7 ⇒ E

8.

# 3x + 3y + 3z

∵ The coefficient of x , y , z have a common factor 3

∵ 3x ÷ 3 = x

∵ 3y ÷ 3 = y

∵ 3z ÷ 3 = z

∴ 3x + 3y + 3z = 3(x + y + z)

∴ The equivalent expression to 3x + 3y + 3z is 3(x + y + z)

∴ 8 ⇒ C

4 0
3 years ago
Central City High's basketball team will be entering the playoffs at the end of their regular season. There will be 3 other team
11Alexandr11 [23.1K]

Answer:

99<x<315

Step-by-step explanation:

4 0
3 years ago
Which value below is included in the solution set for the inequality statement?
mixas84 [53]

You haven't provided any value, but I can tell you the solution set for the inequality.

First of all, expand both sides:

-3x+12 > 6x-6

Add 3x to both sides:

12 > 9x-6

Add 6 to both sides:

18 > 9x

Which is of course equivalent to

9x < 18

Divide both sides by 9

x < 2

So, every number smaller than 2 is part of the solution of this inequality.

7 0
3 years ago
Twenty-two is what percent of 55<br><br> A) 22%<br> B) 40%<br> C) 50%<br> D) 250%
Mashutka [201]

Answer:

twenty two percent of 55 would be the answer B.40 I think.

4 0
3 years ago
How does adding the log together automatically mean that it is a factorial?
andreyandreev [35.5K]

Answer: The answer is given below.

Step-by-step explanation:  We are given an equality involving logarithm and we are to show the implication of L.H.S. to R.H.S.

We will be using the following two properties of logarithm:

(i)~\log_ba=\dfrac{1}{\log_ab},\\\\\\(ii)~log_ab+\log_ac=\log_a(bc).

The proof is as follows:

L.H.S.\\\\\\=\dfrac{1}{\log_2N}+\dfrac{1}{\log_3N}+\dfrac{1}{\log_4N}+\cdots+\dfrac{1}{\log_{100}N}\\\\\\=\log_N2+\logN3+\log_N4+\cdots+\log_N100\\\\=\log_N\{2.3.4...100\}\\\\=\log_N\{1.2.3.4...100\}\\\\=\log_N{100!}\\\\=\dfrac{1}{\log_{100!}N}\\\\=R.H.S.

Hence proved.

6 0
3 years ago
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