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Ivanshal [37]
3 years ago
11

Help please..........

Mathematics
1 answer:
dem82 [27]3 years ago
3 0

Answer:

Step-by-step explanation:

x is greater then -2

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Solve using elimination x-2y=10 and x+3y=5
Zina [86]
Multiply the equation (x-2y=10) by -1
-x+2y=-10
x+3y=5
The x’s cancel out
2y=-10
3y=5
Combine
5y=-5
Divide both sides by 5
y=-1
Plug in y to either original equation
x-2(-1)=10
x+2=10
Subtract 2 from both sides
x=8, y=-1
6 0
3 years ago
find an equation of the line that satisfies the given condition. write the equation in y=mx+b form. passes through points (1,2)
Sindrei [870]

Answer:

find the slope of the points,then simply put the answer as m,then choose any (y) from the points and put it as (c)

3 0
3 years ago
Belinda's family goes through 572 gallons of milk in a year. How many gallons of milk does her family drink every week? (There a
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572 gallons of milk / 52 weeks = 11 gallons of milk / week.

3 0
3 years ago
What is 100 divided by 5, (the answer) x 3 (the answer) x 79 (the answer) - 4
Pie

Answer:

4736

Step-by-step explanation:

1) 100 divided by 5 is 20

2) 20 times 3 is 60

3) 60 times 79 is 4740

4) 4740 subtract from 4 is 4736

5) Answer is 4736

4 0
3 years ago
Read 2 more answers
Write each sum using summation notation<br> 12 + 22 + 32 + 42 + ⋯ + 10000^2
Murljashka [212]

Answer:

50,00,70,000

Step-by-step explanation:

The given sequence is Arithmetic Progression.

Arithmetic Progression is a sequence in which every two neighbor digits have equal distances.

Here first we will find the number of terms

For finding the nth term, we use formula

aₙ = a + (n - 1) d

where, aₙ = value of nth term

a = First term

n = number of term

d = difference

Now, In given sequence: 12, 22, 32, 42, ⋯ , 100002

a = 12, d = 10, n = ? and  aₙ= 100002

∴ 100002  = 12 + (n - 1) × (10)

⇒ 99990 = 10(n - 1)

⇒ n = 10000

Now using the formula of Sum of Arithmetic Sequence,

Sₙ = n÷2[2a + (n - 1)d]

⇒ Sₙ = (10000÷2)[2 × 12 + 9999 × 10]

⇒ Sₙ = 5000 [ 24 + 99990]

⇒ Sₙ = 5000 × 100014 = 50,00,70,000

5 0
3 years ago
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