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3 years ago

6x + 5y = 147 8x + 4y = 156

Mathematics

1 answer:

satela [25.4K]3 years ago

3 0

Answer:

[12, 15]

Step-by-step explanation:

{6x + 5y = 147

{8x + 4y = 156

-¾[8x + 4y = 156]

{6x + 5y = 147

{-6x - 3y = -117 ← New Equation

________________

2y = 30

___ ___

2 2

y = 15 [Plug this back into both equations to get the x-coordinate of 12]; 12 = x

So, using the Elimination Method, what I did here was took the bottom equation and multiplied it by -¾, turning 8x to -6x, canceling out all <em>x-terms</em><em>.</em><em> </em>It does not matter which variable you choose to eliminate, as long as you know what you are doing.

I am joyous to assist you anytime.

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Convert (CAB) 16 to decimal<br><br>need help asap

Mrrafil [7]

Remark

So CAB is a base 16 number? I hope that's correct. I'll treat it that way. If I am incorrect, please leave a note and I'll modify my answer accordingly.

Expand

A = 10 in decimal

B = 11 in decimal

C = 12 in decimal

B is the "units" value in decimal it is 11

A is the "tens" value or 12 place and in decimal you would get 12 * 10 = 120

C is the "hundreds" value or 12^2 place and in decimal you would get 144*12 =1728

Now you need to all all three results together. 11 + 120 + 1728 = 1850

Answer

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6 0

3 years ago

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Colt1911 [192]

Answer:

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Step-by-step explanation:

Because it increases by 3 each time.

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3 years ago

What is 1/2 X (6 X 4)+3 + 2 Please show your work

exis [7]

Answer:

Step-by-step explanation:

Use PEMDAS:

P Parentheses first

E Exponents (ie Powers and Square Roots, etc.)

MD Multiplication and Division (left-to-right)

AS Addition and Subtraction (left-to-right)

$\dfrac{1}{2}\times\underbrace{(6\times4)}_{1}+3+2\\\\=\underbrace{\dfrac{1}{2}\times24}_{2}+3+2\\\\=12+3+2=17$

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3 years ago

Determine the rational zeros for the function f(x)=x^3+7x^2+7x-15

alexandr402 [8]

Answer:

1,-3,-5

Step-by-step explanation:

Given:

f(x)=x^3+7x^2+7x-15

Finding all the possible rational zeros of f(x)

p= ±1,±3,±5,±15 (factors of coefficient of last term)

q=±1(factors of coefficient of leading term)

p/q=±1,±3,±5,±15

Now finding the rational zeros using rational root theorem

f(p/q)

f(1)=1+7+7-15

f(-1)= -1 +7-7-15

= -16

f(3)=27+7(9)+21-15

=96

f(-3)= (-3)^3+7(-3)^2+7(-3)-15

= 0

f(5)=5^3+7(5)^2+7(5)-15

=320

f(-5)=(-5)^3+7(-5)^2+7(-5)-15

f(15)=(15)^3+7(15)^2+7(15)-15

=5040

f(-15)=(-15)^3+7(-15)^2+7(-15)-15

=-1920

Hence the rational roots are 1,-3,-5 !

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3 years ago

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sammy [17]

Answer:

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Step-by-step explanation:

Before we begin, lets define the transformations:

(x, y)

T(a, b) = (x+a, y +b)

Rxaxis = (x, -y)

Ryaxis = (-x, y)

r(180, 0) = (-x, -y)

D.5(1/2x, 1/2y)

R(-90, O) = (-y, x)

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Rx. A(6, 1) B(14,0) C(11,-4)

r180 A(-6,-1) B(-14,0) C(-11,4)

2. Rx D(-12,-6) E(-4,-6) F(-6, -9) G(-10, -9)

Ry D(12,-6) E(4,-6) F(6, -9) G(10, -9)

D.5 D(6,-3) E(2,-3) F(3, -4.5) G(5, -4.5)

3. Rx=1 H(-1,2) I(4, 2) J(4,-2) K(-1,-2)

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r-90 H(-4,1) I(-4, 6) J(0,6) K(0,1)

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