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3 years ago

Express 0.2 as a fraction

Mathematics

2 answers:

professor190 [17]3 years ago

8 0

The 2 is in the tenths place so it's 2/10 or two tenths. : )

tamaranim1 [39]3 years ago

5 0

The simplified answer would be 1/5

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If the sum of two numbers is 90 but their difference is 12 what are the two numbers

emmainna [20.7K]

51 and 39. Have a great day!

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3 years ago

Is the equation below a function? explain d) y2 = x2 + 5

RoseWind [281]

Answer:

Step-by-step explanation:

y^2 = x^2 + 5

Rewriting

y^2 - x^2 = 5

This is a hyperbola

We can use the vertical line test to show that it is not a function since a vertical line passes through more than one point.

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3 years ago

Which expression is the additive inverse of n? A. 1/n B. -n C. -1/n D. -(-n)

Marrrta [24]

The additive inverse of something is whats added to the original to get 0

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4 years ago

Read 2 more answers

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,

$t = 2.60s$

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3 years ago

Write an equation of the line parallel to y = - 3x + 7 y-intercept = 0

polet [3.4K]

A parallel line to y = -3x+7 with a y intercept of 0 is

y = -3x + 0, or you could simply write it as y = -3x

3 0

3 years ago