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3 years ago

10

The sum of two numbers is 73. When the smaller number is subtracted from twice the greater number, the result is 50. Find the nu

mbers. Please use a let statment and an algebraic solution. Thanks! :)

1 answer:

Karolina [17]3 years ago

5 0

Answer:

The big number is 41 and the small number is 32.

Step-by-step explanation:

b + s = 73

2b - s = 50 Add these equations together

3b = 123 Divide both sides by 3

b = 41

Now plug this into the first equation to find s

41 + s = 73

-41 - 41

s = 32

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The product of eight minus a number and seven equals forty when decreased by five. Which equation represents the sentence?

weeeeeb [17]

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So, you are multiplying (8-n), where n = "a number," by 7.

Then, when you decrease (subtract) this product by 5, you get 40.

So the whole thing looks like

7(8-n) - 5 = 40.

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2 years ago

What is the unit cost per bottle if a pack of 6 bottles of juice sells for $4.80

juin [17]

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Step-by-step explanation:

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2 years ago

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3 years ago

Read 2 more answers

Select ALL the correct answers.

IgorC [24]

Answer:

A. 4(x - 4) + 2(3x² + 3x - 20)

C. (11x² + 7x - 55)-(5x² - 3x + 1)

F. (3x² + 5x - 28) + (3x² + 5x - 28)

Step-by-step explanation:

Given:

(3x-7)(2x+8)

= 6x² + 24x - 14x - 56

=6x² + 10x - 56

A. 4(x - 4) + 2(3x² + 3x - 20)

= 4x - 16 + 6x² + 6x - 40

= 6x² + 10x - 56

B. (3x² + 5x - 28) - (2x² + 4x + 28)

= 3x² + 5x - 28 - 2x² - 4x - 28

= x² + x - 56

C. (11x² + 7x - 55)-(5x² - 3x + 1)

= 11x² + 7x - 55 - 5x² + 3x - 1

= 6x² + 10x - 56

D. 4(x - 4) - 2(3x² + 3x - 20)

= 4x - 16 - 6x² - 6x + 40

= - 6x² - 2x + 24

E. (11x² + 7x - 55)-(5x² - 3x + 2)

= 11x² + 7x - 55 - 5x² + 3x - 2

= 11x² - 5x² + 7x + 3x - 55 - 2

= 6x² + 10x - 57

F. (3x² + 5x - 28) + (3x² + 5x - 28)

= 3x² + 5x - 28 + 3x² + 5x - 28

= 6x² + 10x - 56

7 0

2 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:

$\alpha_{1} = -2\cdot \alpha_{3}$

$\alpha_{1} = -2\cdot k$

$\alpha_{2}= -2\cdot \alpha_{3}$

$\alpha_{2} = -3\cdot k$

It is evident that $\alpha_{1}$ and $\alpha_{2}$ are multiples of $\alpha_{3}$ , which means that the set of vector are linearly dependent.

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

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

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

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

The following system of linear equations is obtained:

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

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

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

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

The set of vectors A and C are linearly independent.

4 0

2 years ago