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2 years ago

10 pounds of

Mathematics

1 answer:

skad [1K]2 years ago

7 0

20/10=2.
14/7=2.
2 dollars per pound.

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Hello I need help<br> Find the missing side lengths, leave your answers as radicals in simplest form

Jet001 [13]

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$y = \frac{1}{2} \times \frac{8 \sqrt{3} }{3} \\$

$y = \frac{4 \sqrt{3} }{3} \\$

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$x = \frac{ \sqrt{3} }{2} \times \frac{8 \sqrt{3} }{3} \\$

$x = \frac{8 \times { \sqrt{3} }^{2} }{2 \times 3} \\$

$x = \frac{8 \times 3}{2 \times 3} \\$

$x = \frac{8}{2} \\$

$x = 4$

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4 0

2 years ago

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,

$t = 2.60s$

8 0

3 years ago

What is the value of y?

Slav-nsk [51]

In a triangle all of the angles always add up to 180, so let's set up an equation.

79 + 37 + y = 180
Simplify
116 + y = 180
Subtract 116 from both sides.
y = 64

The value of y is 64.

Hopefully this helps! If you have any more questions or don't understand, please comment or DM me, and I'll get back to you ASAP. :)

6 0

3 years ago

Consider the first five steps of the derivation of the Quadratic Formula.

lara31 [8.8K]

Answer:

Full proof below

Step-by-step explanation:

$\displaystyle ax^2+bx+c=0\\\\ax^2+bx=-c\\\\x^2+\biggr(\frac{b}{a}\biggr)x=-\frac{c}{a}\\\\x^2+\biggr(\frac{b}{a}\biggr)x+\biggr(\frac{b}{2a}\biggr)^2=-\frac{c}{a}+\biggr(\frac{b}{2a}\biggr)^2\\\\x^2+\biggr(\frac{b}{a}\biggr)x+\frac{b^2}{4a^2}=-\frac{c}{a}+\frac{b^2}{4a^2}\\ \\\biggr(x+\frac{b}{2a}\biggr)^2=-\frac{4ac}{4a^2}+\frac{b^2}{4a^2}\\ \\\biggr(x+\frac{b}{2a}\biggr)^2=\frac{b^2-4ac}{4a^2}\\ \\x+\frac{b}{2a}=\pm\frac{\sqrt{b^2-4ac}}{2a}\\ \\x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

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1 year ago

Which equations are true?

saveliy_v [14]

B is true because 0-(-x) = 0 + x = x

5 0

3 years ago