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3 years ago

Which of the following ordered pairs are solutions to the system of equations below?

Mathematics

1 answer:

Masja [62]3 years ago

8 0

The ordered pair (-1, 6) is the solution to the system of equations.

Step-by-step explanation:

Given equations are;

3x+2y=9 Eqn 1

y=4x+10 Eqn 2

Putting value of y from Eqn 2 in Eqn 1

$3x+2(4x+10)=9\\3x+8x+20=9\\11x=9-20\\11x=-11$

Dividing both sides by 11

$\frac{11x}{11}=\frac{-11}{11}\\x=-1$

Putting x=-1 in Eqn 2

$y=4(-1)+10\\y=-4+10\\y=6$

Solution: (-1, 6)

The ordered pair (-1, 6) is the solution to the system of equations.

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Can some please answer these for me!!?

aivan3 [116]

Answer:

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12 - Right Angle Triangle

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3 0

3 years ago

a mixture of peanuts and corn sells for P40 per kilo. The peanuts sell for P42 per kilo while the corn sells for P36 per kilo. h

Lelu [443]

Answer:

The weight of peanuts in the mixture = 8 kg

The weight of corns in the given mixture = 4 kg

Step-by-step explanation:

Let us assume the weight of peanuts in the mixture = x kg

The weight if corns in the given mixture = y kg

Total weight = (x + y) kg

The combined mixture weight = 12 kg

⇒ x + y = 12 ..... (1)

Cost of per kg if mixture = $ 40

So, the cost of (x + y) kg mixture = (x+y) 40 = 40(x+ y) ..... (2)

The cost of 1 kg of peanuts = $ 42

So cost of x kg of peanuts = 42 (x) = 42 x

The cost of 1 kg of corns = $ 36

So cost of y kg of corns = 36 (y) = 36 y

So, the total cost of x kg peanuts + y kg corns = 42 x + 36 y .... (3)

From (1) and (2), we get:

40(x+ y) = 42 x + 36 y

x + y = 12 ⇒ y = 12 -x

Put this in 40(x+ y) = 42 x + 36 y

We get:

40(x+ 12 -x) = 42 x + 36 (12 -x)

480 = 42 x + 432 - 36 x

or, 480 - 432 = 6 x

or, x = 8

⇒ y = 12 -x = 12 - 8 = 4

⇒ y = 4

Hence, the weight of peanuts in the mixture = 8 kg

The weight of corns in the given mixture = 4 kg

5 0

3 years ago

Combine like terms:8y-3y=

Hatshy [7]

Simplitying
8y + - 3y=0

combin like terms : 8y + - 3y= 5y
5y = 0

solving
5y = 0

solving for variable. ' y ' .

move all terms containing y to the left , all other terms to the right

divide each side by ' 5 ' .
y = 0

simplifying
y = 0

6 0

3 years ago

manu took a loan of Rs 5000 for one year at 12%per annum compounded quarterly .How much he has to pay after one year.

Shkiper50 [21]

Answer:

Amount pay after one year for compounded quarterly = Rs 5627.54

Step-by-step explanation:

Given as,

Manu took loan of Rs 5000 , So, Principal = Rs 5000

The rate of interest applied = 12% per annum compounded quarterly

The loan took for period of year = one

Now from the compounded method :

For compounded quarterly

Amount = principal $(1 + \frac{Rate}{4\times 100})^{4\times Time}$

Or, Amount = Rs 5000 $(1+ \frac{12}{4\times 100})^{4}$

Or, Amount = 5000 $(\frac{103}{100})^{4}$

Or, Amount = 5000 × 1.1255

∴ Amount = Rs 5627.54

Hence , The amount which Manu pay after one year at 12% per annum compounded quarterly is Rs 5627.54 Answer

5 0

3 years ago

Can someone help me?

Lana71 [14]

Answer:

I'm not sure from my answer but I'm going to try

A = { a^2 - 4 = 0 }

A = { a^2 = 0+ 4 }

A = { a^2 = 4}

Take root of both sides

A = {a = 2}

A = {2}

B = { -2, -1, 0, 1, 2}

C = { 1, 2, 3, 4, 5, 6,7}

i) A\B

= {2} \ {-2,-1,0,1,2}

= ⌀ ( "null set" )

ii) (A\B) \C

= ( {2} \ {-2,-1,0,1,2} ) \ C

= ⌀ \ C

= ⌀ \ { 1, 2, 3, 4, 5, 6,7}

= ⌀ ( "null set" )

iii) (A ∪ B) \ C

= ( { 2 } ∪ { -2, -1, 0, 1, 2} ) \ C

= { -2, -1, 0, 1, 2} \C

= { -2, -1, 0, 1, 2} \ { 1, 2, 3, 4, 5, 6,7}

= { -2, -1, 0 }

iv) ( A\B) ∩ ( A\C)

= ( { 2 } \ {-2,-1,0,1,2} ) ∩ ( { 2 } \ { { 1, 2, 3, 4, 5, 6,7} )

= ⌀ ∩ ⌀

= ⌀

v) ( A ∩ B ) \ C

= ( { 2 } ∩ { -2, -1, 0, 1, 2 } ) \ C

= { 2 } \ { 1, 2, 3, 4, 5, 6, 7 }

= ⌀ "null set"

7 0

3 years ago