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Damm [24]
3 years ago
9

How to solve a\10=-10

Mathematics
1 answer:
koban [17]3 years ago
4 0

Answer:

a = - 100

Step-by-step explanation:

Given

\frac{a}{10} = - 10

Multiply both sides by 10 to clear the fraction

a = 10 × - 10 = - 100

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Steven's savings account switches from compounding interest annually to quarterly. His account earns 3% interest yearly. Steven
V125BC [204]

Answer:

$0.95 more.

Step-by-step explanation:

The principal of $500, when invested at APR of 3% for 5 years compounded annually will become

S_1 = 500(1 + \frac{3}{100})^{5} = 579.64 dollars.

Again, the principal of $500, when invested at APR of 3% for 5 years compounded quarterly will become

S_2 = 500(1 + \frac{3}{100 \times 4})^{5 \times 4} = 580.59 dollars.

Therefore, Steven will have $(580.59 - 579.64) = $0.95 more money in his account due to switching from annually to quarterly compounding. (Answer)

8 0
3 years ago
Which of the following lines has a negative slope?
Furkat [3]

Answer:

The second option, or the one attached.

Step-by-step explanation:

       This line has a negative slope because the line is going "down the stairs" from the top-left to the bottom-right.

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2 years ago
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Nata [24]

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3 years ago
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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.

4 0
3 years ago
Josh works for The Milkshake Diner. The company wants to know which milkshake flavor is the most popular. Today, he surveyed
AlekseyPX

its b i think please vote me i need points

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