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bagirrra123 [75]

3 years ago

12

How are you expose to solve 7n-7n=1

1 answer:

eduard3 years ago

6 0

You just subtract normally because they both have the same variable. I hope you find this helpful.

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Answer please 1st and second

defon

1/9 x 12 is 4/3 or 1.333 repeated

8 0

3 years ago

Read 2 more answers

13 and 5/8 plus 7/8 please answer quickly

Salsk061 [2.6K]

13 5/8 + 7/8 = 14 1/2

5/8 + 7/8 = 1 4/8 = 1 1/2

1 1/2 + 13 = 14 1/2

8 0

3 years ago

How do you do 1/4 (16x+20)-7x

Ira Lisetskai [31]

Distribute the 1/4:
1/4(16x) + 1/4(20) - 7x

Do the multiplication:
1/4(16x) = 16x/4 = 4x
1/4 (20) = 20/4 = 5

Now you have:
4x + 5 - 7x = 0

Combine like terms (4x and -7x):
-3x + 5 = 0

Subtract 5 from both sides:
-3x = -5

Divide by -3:
x = -5/-3 = 5/3

Answer:
x= 5/3

6 0

3 years ago

Classify ABC by its sides. Then determine whether it is a right triangle.

m_a_m_a [10]

Answer:

∴Given Δ ABC is not a right-angle triangle

a= AB = √45 = 3√5

b = BC = 12

c = AC = √45 = 3√5

Step-by-step explanation:

Given vertices are A(3,3) and B(6,9)

$AB = \sqrt{x_{2}-x_{1} )^{2} +(y_{2}-y_{1} )^{2} }$

AB = $\sqrt{(9-3)^{2}+(6-3)^{2} } = \sqrt{6^{2}+3^{2} } =\sqrt{45}$

Given vertices are B(6,9) and C( 6,-3)

$B C = \sqrt{x_{2}-x_{1} )^{2} +(y_{2}-y_{1} )^{2} }$

= $\sqrt{(-3-9)^{2}+(6-6)^{2} } =\sqrt{12^{2} } = 12$

BC = 12

Given vertices are A(3,3) and C( 6,-3)

$AC = \sqrt{(6-3)^{2}+(-3-3)^{2} } = \sqrt{9+36} = \sqrt{45}$

AC² = AB²+BC²

45 = 45+144

45 ≠ 189

∴Given Δ ABC is not a right angle triangle

5 0

3 years ago

I really need help, I haven't been to school In a while and I kinda forgot what to do. So If anyone is kind enough, please help

marissa [1.9K]

Answer:

P = 20x + 6y + 10

Step-by-step explanation:

The perimeter (P) of the figure is the sum of the sides , that is

P = 2x - y + 3x + 5y + 12 + 6x + 9x + 2y - 2 ← collect like terms

= 20x + 6y + 10

3 0

3 years ago