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ValentinkaMS [17]
3 years ago
5

Hey ya'll don't scroll please, just look at the attachment and answer every single question please- thank you very much.

Mathematics
1 answer:
Gelneren [198K]3 years ago
7 0

Answer:

A) pop? If you meant most

B.)30

C.) 230

Step-by-step explanation:

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Write an equation in point slope form for the line that passes through the given point and is perpendicular to the graph of the
USPshnik [31]

Answer:

y=-3x-8

Step-by-step explanation:

y=1/3x+9

slope is 1/3

perpendicular slope = m

(m)(1/3) = -1

m = -3

Point slope form

(y+2)/(x+2)=-3

y+2=-3x-6

y=-3x-8

4 0
3 years ago
How to solve (8x-30)+(6x+10)=180
lozanna [386]
Add the like terms. So I'll add the variables. 8x+6x is 14x. Now add whole numbers. 30+10 is 40. Our new expression we'll be 14x+40=180. Subtract 40 from 180. 180-40=140. Now we have the expression 14x=140. 140 divided by 14 is 10. So x is 10.
7 0
3 years ago
How would you explain the proper use of distributive property ? ​
Maksim231197 [3]
Start with the word “distribute” meaning to pass equally. Now distributive property means to distribute equally. For example 2(4x+5) you wanna distribute the 2 amount the parentheses so that it’s”distributed evenly. 2(4x+5) is now 8x+10
3 0
3 years ago
Jeremiah Holmes
olga nikolaevna [1]

Answer:

- 508

Step-by-step explanation:

The n th term of an arithmetic sequence is

a_{n} = a₁ + (n - 1)d

where a₁ is the first term and d the common difference

Here a₁ = - 28 and d = - 34 - (- 28) = - 34 + 28 = - 6, thus

a_{81} = - 28 + (- 6 × 80) = - 28 - 480 = - 508

6 0
3 years ago
The volume ( v) of a gas in a container at a constant temperature varies inversely as the pressure (p). If the volume of a gas i
andre [41]
The answer is 12.8 cm³

<span>The volume ( v) of a gas in a container at a constant temperature varies inversely as the pressure (p):
v = k/p
p * v = k  (k - constant)

p1v1 = k
p2v2 = k
So, p1v1 = p2v2

v1 = 24 cm</span>³
p1 = 16 lb
v2 = ?
p2 = 30 lb

24 * 16 = x * 30
384 = x * 30
x = 384 / 30
x = 12.8 cm³
3 0
3 years ago
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