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valentinak56 [21]
2 years ago
10

Which is the x-value of the solution of the system of equations? y=3x^2+15 y=5x^2-17

Mathematics
1 answer:
Readme [11.4K]2 years ago
8 0

Answer:

C. x = ±4

Explanation:

<u>Given two equation's</u>:

  1. y = 3x² + 15
  2. y = 5x² - 17

Solve them simultaneously:

 y       =       y

3x² + 15 = 5x² - 17

5x² - 3x² = 15 + 17

2x² = 32

x² = 16

x = ±√16

x = ±4

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A carton has a length of 2 2/3, width of 1 1/3, and height of 1 1/2 feet. What is the volume of the carton?
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3 years ago
A parallelogram has a base of 4.5 cm and an area of 9.495 cm². Tania wrote the equation 4.5x = 9.495 to represent this situation
Artist 52 [7]

Answer:

The height of the parallelogram is 2.11 cm.

Step-by-step explanation:

Area of the parallelogram is equal to multiplication of base and height.

Given:

Parallelogram has base of 4.5 cm.

Are of the parallelogram is 9.495 cm².

Equation is 4.5x=9.495

Calculation:

(a)

Are of the parallelogram is the product of base length and height of the parallelogram.

Area of the parallelogram is expressed as follow:

A=lh          

Substitute 9.495 cm² for A and 4.5 cm for l in above equation as follows:

9.495=4.5h             …… (1)

Now relate the equation (1) and given equation. So, here x is nothing but the height of the parallelogram.

(b)

From equation (1), height of the parallelogram is calculated as follows:

9.495=4.5h

h=\frac{9.495}{4.5}

h=2.11 cm

Thus, the height of the parallelogram is 2.11 cm.

5 0
3 years ago
Verify that the function satisfies the three hypotheses of Rolle's Theorem on the given interval. Then find all numbers c that s
WINSTONCH [101]

Answer:  \bold{c=\dfrac{1+\sqrt{19}}{3}\approx1.8}

<u>Step-by-step explanation:</u>

There are 3 conditions that must be satisfied:

  1. f(x) is continuous on the given interval
  2. f(x) is differentiable
  3. f(a) = f(b)

If ALL of those conditions are satisfied, then there exists a value "c" such that c lies between a and b and f'(c) = 0.

f(x) = x³ - x² - 6x + 2     [0, 3]

1. There are no restrictions on x so the function is continuous \checkmark

2. f'(x) = 3x² - 2x - 6 so the function is differentiable \checkmark

3. f(0) =  0³ - 0² - 6(0) + 2 = 2

   f(3) =  3³ - 3² - 6(3) + 2  = 2

   f(0) = f(3) \checkmark

f'(x) = 3x² - 2x - 6 = 0

This is not factorable so you need to use the quadratic formula:

x=\dfrac{-(-2)\pm \sqrt{(-2)^2-4(3)(-6)}}{2(3)}\\\\\\.\quad =\dfrac{2\pm \sqrt{4+72}}{2(3)}\\\\\\.\quad =\dfrac{2\pm \sqrt{76}}{2(3)}\\\\\\.\quad =\dfrac{2\pm 2\sqrt{19}}{2(3)}\\\\\\.\quad =\dfrac{1\pm \sqrt{19}}{3}\\\\\\.\quad \approx\dfrac{1+4.4}{3}\quad and\quad \dfrac{1-4.4}{3}\\\\\\.\quad \approx1.8\qquad and\quad -1.1

Only one of these values (1.8) is between 0 and 3.

5 0
3 years ago
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