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3 years ago

How does a credit score affects creditworthiness and the cost of credit?

1 answer:

5 0

Here is the answer. Credit score affects credit worthiness and the cost of credit by <span>short term loans, swipe fees, interest use networking banks. Credit score is the analysis of the person's credit files, and this is the basis of his credit worthiness. Hope this answers your question.</span>

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Match the following reasons with the statements in the given proof. Prove that if two angles are complementary to the same angle

yulyashka [42]

1. --> d

2. --> b

3. --> a

4. --> c

1 is d because that is the only piece of information that sounds like a given.

2 is b because we know that from the given, m∠1 is complementary to m∠2 and in the first problem, m∠1 + m∠2 = 90, they are added together. The definition of complementary angles is "Two angles with measures that, when added together, equal 90 degrees". Same thing with m∠3 and m∠2.

3 is a because since both of the problems equal 90 degrees, you can just take away the 90 and put an equal sign in between the two problems because they equal the same thing.

4 is c because you are subtracting m∠2 from the problems and ending up with just m∠1 = m∠3.

Hope this helps! :)

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3 years ago

Read 2 more answers

To make subtracting integers easier you can change all subtraction problems to

Irina18 [472]

Answer:

ssssss

Step-by-step explanation:

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4 years ago

What is the question 15%of258

Readme [11.4K]

The answer would be 219.3 (:
Please mark me the brainliest ty

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3 years ago

Please help!!!!!!!!!

Maru [420]

Answer: A

Step-by-step explanation:

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3 years ago

A function $f$ has a horizontal asymptote of $y = -4,$ a vertical asymptote of $x = 3,$ and an $x$-intercept at $(1,0).$ Part (a

Masja [62]

1. $f$ has a horizontal asymptote at $y=-4$

This means that

$\displaystyle\lim_{x\to\pm\infty}f(x)-(-4)=0$

(for at least one of these limits)

2. $f$ has a vertical asymptote at $x=3$

This means that $f$ has a non-removable discontinuity at $x=3$ . Since $f$ is some rational function, there must be a factor of $x-3$ in its denominator.

3. $f$ has an $x$ -intercept at (1, 0)

This means $f(1)=0$ .

(a) With

$f(x)=\dfrac{ax+b}{x+c}$

the second point above suggests $c=-3$ . The first point tells us that

$\displaystyle\lim_{x\to\pm\infty}\frac{ax+b}{x-3}+4=0=\lim_{x\to\pm\infty}\frac{ax+b+4x-3}{x-3}=0$

In order for the limit to be 0, the denominator's degree should exceed the numerator's degree; the only way for this to happen is if $a=-4$ so that the linear terms vanish.