The Cramer's Rule states that, in a system of equation in two unknowns, in which determinants D is not 0, X=_and Y=_.

Mathematics

1 answer:

yan [13]3 years ago

5 0

Answer:

Step-by-step explanation:

$x=\frac{D_{1}}{D} \\y=\frac{D_{2}}{D}$

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with reference to a universal set, the inclusion of a subset in another is relation which is a) symmetric only b) transitive onl

Vlad [161]

Answer:

Step-by-step explanation:

I think it is D since a universal set would only be {1,2,3,4,5,6,...} and the subset would not satisfy any of the conditions of symmetric, transitive, and reflexive.

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3 years ago

Which ordered pair is a solution of the equation?<br> y=4x−7

vampirchik [111]

Answer:

Unknown

Step-by-step explanation:

Hi! so you didn't add the ordered pair options but I will do my best :) so if you see the ordered pair (0.7) that could definitely be a solution because using the slope intercept form equation you gave I know that's our y Intercept. the slope is 4/1 so moving up 4 and to the right 1 or down 4 and to the left 1 on a graph could show you more points! If you don't want to draw it you could use Desmos and just put this equation in or give me the ordered pairs and I will give you your answer!

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2 years ago

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Zigmanuir [339]

Answer:

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Step-by-step explanation:

4-6 3-5

------- , ---------

2 2

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3 years ago

At an effective annual interest rate of i > 0, each of the following two sets of payments has present value K: (i) A payment

IgorLugansk [536]

Answer:

The present value of K is, $K=251.35$

Step-by-step explanation:

First of all, we need to construct an equation system, so

$(1)K=\frac{169}{(1+i)} +\frac{169}{(1+i)^{2}}$

$(2)K=\frac{225}{(1+i)^{2}} +\frac{225}{(1+i)^{4}}$

Then we equalize both of them so we can find $i$

$(3)\frac{169}{(1+i)} +\frac{169}{(1+i)^{2}}=\frac{225}{(1+i)^{2}} +\frac{225}{(1+i)^{4}}$

To solve it we can multiply $(3)*(1+i)^{4}$ to obtain $(1+i)^{4}*(\frac{169}{(1+i)} +\frac{169}{(1+i)^{2}}=\frac{225}{(1+i)^{2}} +\frac{225}{(1+i)^{4}})$ , then we have $225(1+i)^{2}+225=169(1+i)^{3}+169(1+i)^{2}$ .