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3 years ago

20000000000×3-100000

Mathematics

1 answer:

Blababa [14]3 years ago

4 0

Answer:

59999900000

Step-by-step explanation:

20000000000 × 3 = 60000000000

60000000000 - 100000 = 59999900000

can I have brainllest

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bixtya [17]

The answer is C. Good luck

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3 years ago

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The sets F and H are given below.

Vinil7 [7]

Answer:

F ∪ H = {c, d, e, f, g, h}

F ∩ H = { }

Step-by-step explanation:

The union is the list of elements that are in either of the two sets.

F ∪ H = {c, d, e, f, g, h}

The intersection is the list of only those elements that appear in both sets. (There are none.)

F ∩ H = { } . . . . the empty set

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3 years ago

Plz help... must show steps....will pick brainiest Solve, finding all solutions in [0,2pi)

Triss [41]

A graphing calculator shows solutions to be
.. x = {π/4, 2.034, 5π/4, 5.176}

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Asked and answered elsewhere.
brainly.com/question/8691400

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3 years ago

1. Kaleb worked at a toy store during his summer vacation from

aleksandr82 [10.1K]

Answer:

1. $10.56

Step-by-step explanation:

950.40÷90=10.56

7 0

3 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

20000000000×3-100000​

20000000000×3-100000