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3 years ago

What is 11.4 rounded to the nearest tenth

Mathematics

1 answer:

mixas84 [53]3 years ago

6 0

11.4 should be the answer
if you mean tens then it is 10

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2/3n + 5 greater than 12

Tanzania [10]

2/3n + 5 > 12
2/3n > 12 - 5
2/3n > 7
n > 7 * 3/2
n > 21/2 or 10 1/2

6 0

3 years ago

Read 2 more answers

Choose all answers that apply:<br> a - -1.5 < x < -0.5<br> b- 0 c- 3.5 d- none of the above

Darya [45]

Answer:

None of the above.

Step-by-step explanation:

The interval where h is increasing is -1.5 < x < 1.5. Answers a, b, and c do not match.

3 0

3 years ago

Sam has $440 in $20, $10, and $5 dollar bills. The number of twenties, tens and fives are consecutive integers in this order. Ho

Ludmilka [50]

Answer: There are 12 bills for $20, 13 bills for $10 and 14 bills for $5.

Step-by-step explanation:

Let the number of twenties be x

Let the number of tens be x+1

Let the number of fives be x+2

Amount from twenties will be given as

$20\times x=20x$

Amount from tens will be given as

$10\times (x+1)=10x+10$

Amount from fives will be given as

$5\times (x+2)=5x+10$

According to question,

$20x+10x+10+5x+10=440\\\\35x+20=440\\\\35x=440-20\\\\35x=420\\\\x=\frac{420}{35}\\\\x=12$

So, there are 12 bills for $20, 13 bills for $10 and 14 bills for $5.

6 0

3 years ago

Find the missing side or angle.

jolli1 [7]

Answer:

3.7

Step-by-step explanation:

Acellus

8 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: