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4 years ago

How is the following decimal number read?

Mathematics

2 answers:

harkovskaia [24]4 years ago

7 0

Answer:

C. Twenty-eight and nine tenths.

Step-by-step explanation:

We have been given a decimal number $28.9$ . We are asked to choose the correct option that represents how to read our given decimal number.

Upon looking at our given decimal number we can see that it has one digit at tenth place that is 9, which represents that our number has 9 tenths.

The two digits before decimal are 2 (tens place) and 8 (ones place). These digits make 28.

Therefore, the given decimal number could be read as twenty-eight and nine tenths and option C is the correct choice.

forsale [732]4 years ago

3 0

Answer:
C) twenty-eight and nine tenths

Explanation:
We have two parts here, the whole number and the decimal part.

1- for the whole number, it is read normally as counting numbers. Therefore, 28 is read as twenty-eight
2- for the decimal part, it is read based on the place-value of the number. We know that the first number after the decimal point is in the tenth position. This means that 0.9 is read as nine tenths

Combining the two, we would conclude that 28.9 is read as twenty-eight and nine tenths

Hope this helps :)

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3 years ago

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Step-by-step explanation:

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3 years ago

Given vectors u = (−1, 2, 3) and v = (3, 4, 2) in R 3 , consider the linear span: Span{u, v} := {αu + βv: α, β ∈ R}. Are the vec

julia-pushkina [17]

Answer:

$(2,6,6) \not \in \text{Span}(u,v)$

$(-9,-2,5)\in \text{Span}(u,v)$

Step-by-step explanation:

Let $b=(b_1,b_2,b_3) \in \mathbb{R}^3$ . We have that $b\in \text{Span}\{u,v\}$ if and only if we can find scalars $\alpha,\beta \in \mathbb{R}$ such that $\alpha u + \beta v = b$ . This can be translated to the following equations:

1. $-\alpha + 3 \beta = b_1$

2. $2\alpha+4 \beta = b_2$

3. $3 \alpha +2 \beta = b_3$

Which is a system of 3 equations a 2 variables. We can take two of this equations, find the solutions for $\alpha,\beta$ and check if the third equationd is fulfilled.

Case (2,6,6)

Using equations 1 and 2 we get

$-\alpha + 3 \beta = 2$

$2\alpha+4 \beta = 6$

whose unique solutions are $\alpha =1 = \beta$ , but note that for this values, the third equation doesn't hold (3+2 = 5 $\neq$ 6). So this vector is not in the generated space of u and v.

Case (-9,-2,5)

Using equations 1 and 2 we get

$-\alpha + 3 \beta = -9$

$2\alpha+4 \beta = -2$

whose unique solutions are $\alpha=3, \beta=-2$ . Note that in this case, the third equation holds, since 3(3)+2(-2)=5. So this vector is in the generated space of u and v.