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s344n2d4d5 [400]
3 years ago
10

3x-5y=60 4x+5y=-4 what is the system of linear equation

Mathematics
1 answer:
Sergeu [11.5K]3 years ago
4 0

(3x-5y=60)

4x+5y=-4

3(8)-5(7.2)=60

4(8)+5(-7.2)=-4

7x=56

÷7. ÷7

x=8

3(8)-5y=60

4(8)+5y=-4

32+5y=-4

-32. -32

5y=-36

÷5. ÷5

y=7.2

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The weight, in pounds, of a newborn baby tt months after birth can be modeled by the function W(t)=1.25t+6.W(t)=1.25t+6. What is
ollegr [7]

Answer:

6

Step-by-step explanation:

GIVEN: The weight, in pounds, of a newborn baby t months after birth can be modeled by the function W(t)=1.25t+6.

TO FIND: What is the y-intercept of the function and what is its interpretation in the context of the problem.

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if t=0

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Hence y-intercept is 6

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2 years ago
Erik and Caleb were trying to solve the equation: 0=(3x+2)(x-4) Erik said, "The right-hand side is factored, so I'll use the zer
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Answer:

C) Both

Step-by-step explanation:

The given equation is:

0=(3x+2)(x-4)

To solve the given equation, we can use the Zero Product Property according to which if the product <em>A.B = 0</em>, then either A = 0 OR B = 0.

Using this property:

(3x+2) = 0 \Rightarrow \bold{x = -\frac{2}{3}}\\(x-4) = 0 \Rightarrow \bold{x = 4}

So, Erik's solution strategy would work.

Now, let us discuss about Caleb's solution strategy:

Multiply (3x+2)(x-4) i.e. 3x^2-12x+2x-8 = 3x^2-10x-8

So, the equation becomes:

0=3x^2-10x-8

Comparing this equation to standard quadratic equation:

ax^2+bx+c=0

a = 3, b = -10, c = -8

So, this can be solved using the quadratic formula.

x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}

x=\dfrac{-(-10)\pm\sqrt{(-10)^2-4\times3 \times (-8)}}{2\times 3}\\x=\dfrac{-(-10)\pm\sqrt{196}}{6}\\x=\dfrac{10\pm14}{6} \\\Rightarrow x= 4, -\dfrac{2}{3}

The answer is same from both the approaches.

So, the correct answer is:

C) Both

3 0
3 years ago
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