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1 year ago

Answer the equation

Mathematics

1 answer:

ivanzaharov [21]1 year ago

4 0

Answer:

x = -2, y = -4

coordinate: (-2, -4)

Step-by-step explanation:

Given system of equations:

a) 4x - 7y = 20

b) x - 3y = 10

1. Make x the subject in the second equation:

⇒ x - 3y = 10 [add 3y to both sides]

⇒ x - 3y + 3y = 10 + 3y

⇒ x = 3y + 10

2. Substitute the given value of x into first equation:

⇒ 4x - 7y = 20

⇒ 4(3y + 10) - 7y = 20 [distribute 4 through the parentheses]

⇒12y + 40 - 7y = 20 [combine like terms]

⇒ 5y + 40 = 20 [subtract 40 from both sides]

⇒ 5y - 40 - 40 = 20 - 40

⇒ 5y = -20 [divide both sides by 5]

⇒ 5y ÷ 5 = -20 ÷ 5

⇒ y = -4

3. Find the value of x by substituting the given value of y into x = 3y + 10:

⇒ x = 3y + 10

⇒ x = 3(-4) + 10 [multiply]

⇒ x = -12 + 10 [add]

⇒ x = -2

4. Check your work:

a) 4x - 7y = 20

⇒ 4(-2) - 7(-4) = 20

⇒ -8 + 28 = 20

⇒ 20 = 20 ✔

b) x - 3y = 10

⇒ -2 - 3(-4) = 10

⇒ -2 + 12 = 10

⇒ 10 = 10 ✔

This system of equations has a solution at (-2, -4).

Learn more here:

brainly.com/question/27849342

brainly.com/question/27728118

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A juggler tosses a ball into the air . The balls height, h and time t seconds can be represented by the equation h(t)= -16t^2+40

malfutka [58]

PART A

The given equation is

$h(t) = - 16 {t}^{2} + 40t + 4$

In order to find the maximum height, we write the function in the vertex form.

We factor -16 out of the first two terms to get,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t) + 4$

We add and subtract

$- 16(- \frac{5}{4} )^{2}$

to get,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t) + - 16( - \frac{5}{4})^{2} - -16( - \frac{5}{4})^{2} + 4$

We again factor -16 out of the first two terms to get,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t + ( - \frac{5}{4})^{2} ) - -16( - \frac{5}{4})^{2} + 4$

This implies that,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t + ( - \frac{5}{4}) ^{2} ) + 16( \frac{25}{16}) + 4$

The quadratic trinomial above is a perfect square.

$h(t) = - 16 ( t- \frac{5}{4}) ^{2} +25+ 4$

This finally simplifies to,

$h(t) = - 16 ( t- \frac{5}{4}) ^{2} +29$

The vertex of this function is

$V( \frac{5}{4} ,29)$

The y-value of the vertex is the maximum value.

Therefore the maximum value is,

$29$

PART B

When the ball hits the ground,

$h(t) = 0$

This implies that,

$- 16 ( t- \frac{5}{4}) ^{2} +29 = 0$

We add -29 to both sides to get,

$- 16 ( t- \frac{5}{4}) ^{2} = - 29$

This implies that,

$( t- \frac{5}{4}) ^{2} = \frac{29}{16}$

$t- \frac{5}{4} = \pm \sqrt{ \frac{29}{16} }$

$t = \frac{5}{4} \pm \frac{ \sqrt{29} }{4}$

$t = \frac{ 5 + \sqrt{29} }{4} = 2.60$

or

$t = \frac{ 5 - \sqrt{29} }{4} = - 0.10$

Since time cannot be negative, we discard the negative value and pick,

$t = 2.60s$