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3 years ago

Help write them in standard form 20points do all

Mathematics

1 answer:

Gemiola [76]3 years ago

8 0

1. 437,000,003,000,000
2. 53,000,200
3. 797,619,000,000,132
4. 11,006,007
5. 84,000,037,800

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To the person who makes a graph and doesn't copy from another question

allochka39001 [22]

Answer:

H=9;8)

B=(5;4) (ball)

R=(7;0) (hit point)

B'=symetric of B axis perpendicular of x in R

B'=(7+(7-5);4)=(9;4)

Equation BR: y-4=(0-4)/(7-5)(x-5)==>y=-2x+14

Equation RB': y-4=(4-0)/(9-7)(x-9)==>y=2x-14

Is H a point of RB'? y=2x-14 : 8=? 2*9-14==>8=?4 No!

you will not make your putt

Step-by-step explanation:

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2 years ago

The temperatures (in degrees Fahrenheit) in Long Island recorded by the weather bureau over a week were 42, 49, 53, 55, 50, 47,

Ghella [55]

The IQR or interquartile range is the difference between Q1 and Q3, also know as the upper and lower quartiles.

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3 years ago

Read 2 more answers

What is 5/6 + (-4/9) - (-2)

timofeeve [1]

Answer:

2 7/18.

common denominator

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3 years ago

1/2x+5=1/3x what do you multiply the equation by in order to get rid of the fractions in the problem

earnstyle [38]

Answer:

we should multiply by 6 to get rid of the fractions in problem

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3 years ago

Determine which of the sets of vectors is linearly independent. A: The set where p1(t) = 1, p2(t) = t2, p3(t) = 3 + 3t B: The se

defon

Answer:

The set of vectors A and C are linearly independent.

Step-by-step explanation:

A set of vector is linearly independent if and only if the linear combination of these vector can only be equalised to zero only if all coefficients are zeroes. Let is evaluate each set algraically:

$p_{1}(t) = 1$ , $p_{2}(t)= t^{2}$ and $p_{3}(t) = 3 + 3\cdot t$ :

$\alpha_{1}\cdot p_{1}(t) + \alpha_{2}\cdot p_{2}(t) + \alpha_{3}\cdot p_{3}(t) = 0$

$\alpha_{1}\cdot 1 + \alpha_{2}\cdot t^{2} + \alpha_{3}\cdot (3 +3\cdot t) = 0$

$(\alpha_{1}+3\cdot \alpha_{3})\cdot 1 + \alpha_{2}\cdot t^{2} + \alpha_{3}\cdot t = 0$

The following system of linear equations is obtained:

$\alpha_{1} + 3\cdot \alpha_{3} = 0$

$\alpha_{2} = 0$

$\alpha_{3} = 0$

Whose solution is $\alpha_{1} = \alpha_{2} = \alpha_{3} = 0$ , which means that the set of vectors is linearly independent.

$p_{1}(t) = t$ , $p_{2}(t) = t^{2}$ and $p_{3}(t) = 2\cdot t + 3\cdot t^{2}$

$\alpha_{1}\cdot p_{1}(t) + \alpha_{2}\cdot p_{2}(t) + \alpha_{3}\cdot p_{3}(t) = 0$

$\alpha_{1}\cdot t + \alpha_{2}\cdot t^{2} + \alpha_{3}\cdot (2\cdot t + 3\cdot t^{2})=0$

$(\alpha_{1}+2\cdot \alpha_{3})\cdot t + (\alpha_{2}+3\cdot \alpha_{3})\cdot t^{2} = 0$

The following system of linear equations is obtained:

$\alpha_{1}+2\cdot \alpha_{3} = 0$

$\alpha_{2}+3\cdot \alpha_{3} = 0$

Since the number of variables is greater than the number of equations, let suppose that $\alpha_{3} = k$ , where $k\in\mathbb{R}$ . Then, the following relationships are consequently found: