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Andru [333]
4 years ago
11

Find the 72nd term of the arithmetic sequence 11, 7, 3, ...

Mathematics
1 answer:
Katen [24]4 years ago
3 0

Answer:

-273

Step-by-step explanation:

find the common difference:

d = 7 - 11 = - 4

a_1 = 11

a_72 =  (72 - 1)* (-4)  +  a_1

a_72 =  (71)* (-4)  +  11

a_72 =   -273

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Perform the specified operations on the polynomials: Subtract 2x2−3xy+4y2from 3x2−5xy−6y2
tankabanditka [31]

Answer:

Solution to the given expression is (x^2 -10y^2 -2xy)

Step-by-step explanation:

Here, the given expression are:

P(x) = 2x^2- 3xy+4y^2\\Q(x) = 3x^2 - 5xy-6y^2

To Solve for : Q(x)  - P(x)

Now substitute the value of P(x) and Q(x) , we get:

Q(x) - P(x) = P(x) = (3x^2 - 5xy-6y^2) - (2x^2- 3xy+4y^2)\\= 3x^2 - 5xy-6y^2 - 2x^2+3xy-4y^2\\=x^2-10y^2-2xy\\\implies Q(x) - P(x) = (x^2-10y^2-2xy)

Hence solution to the given expression is (x^2 -10y^2 -2xy)

8 0
4 years ago
If 0.5x-3=2-0.75x then 2x =?
Anvisha [2.4K]

Answer:

8

Step-by-step explanation:

Step 1:

0.5x - 3 = 2 - 0.75x  Equation

Step 2:

1.25x - 3 = 2     Add 0.75x on both sides

Step 3:

1.25x = 5     Add 3 on both sides

Step 4:

5 ÷ 1.25  Divide

Step 5:

x = 4  Solution to x

Step 6:

2 × 4    Multiply to find answer of 2x

Answer:

2x = 8

Hope This Helps :)

5 0
3 years ago
X = t - 3, y equals two divided by quantity t plus five
nekit [7.7K]
X = t - 3 is also -t = -x - 3 or t=x+3

Substitute it in

y=2/t+5
y=2/(x+3)+5

Let y=0 to solve for x

0=2/(x+3)+5
-5= 2/(x+3)
-10=x+3
-7=x

Now solve for t

t = x+3
t = (-7)+3
t = -4

Now we can solve for y

y = 2/(-4)+5
y = -0.5+5
y = 4.5

To check

4.5 = 2/(-4) + 5
4.5 = -0.5 + 5
4.5 = 4.5

-7 = -4 - 3
-7 = -7

Therefore
x = -7
y = 4.5
t = -4
5 0
3 years ago
25=9x+74-7<br> plz help brainz too
RideAnS [48]

Answer:

-14/3

Step-by-step explanation:

25=9x+74-7

25=9x+67

-42=9x

-42/9=9x/9

-42/9=x

-14/3=x

4 0
3 years ago
The length of a rectangle plus its width is 28
Alex17521 [72]

Answer:

The length and width of the rectangle is 20 cm and 8 cm.

Step-by-step explanation:

3 0
3 years ago
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