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3 years ago

What common fraction is equivalent to 7.5%?

Mathematics

1 answer:

kykrilka [37]3 years ago

7 0

Answer:

3/40

Step-by-step explanation:

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Solve the following system of equations by elimination. What is the value of x? x = 1.5 x = -1.5 x = 8 x = -8

Andru [333]

Using the elimination method, the value of x in the system of equations is calculated as: 8.

<h3>How to Solve a System of Equations by Elimination?</h3>

To solve a system of equations given using the elimination method, do the following:

Multiply 2x - 5y = 1 by 2 and multiply -3x + 2y = -18 by 5 to get the following:

4x - 10y = 2 --> eqn. 1

-15x + 10y = -90 --> eqn. 2

Add

-11x = -88

Divide both sides by -11

x = 8

Learn more about system of equations on:

brainly.com/question/14323743

#SPJ1

8 0

2 years ago

The graph of a function is shown below. Which ordered pair best represents the location of the y intercept

Wittaler [7]

When a question is asking you for the y-intercept based on a graph, all you have to do is look to see at what point does the function cross/intersect the y-axis. The x-value will always be zero in a y-intercept. By looking at the graph you see that the function crosses the y-axis at the point (0, -1), which is your answer.

Answer: (0, -1)

8 0

3 years ago

Expand and simplify 2(5x - 1) – (2x - 5)

horrorfan [7]

Answer:

2(5x - 1) – (2x - 5)

=2(5x−1)−2x+5

=10x−2−2x+5

=(10x−2x)+(−2+5)

=8x+3

Step-by-step explanation:

3 0

3 years ago

The expression P(4, 3) is equal to:<br> -3!<br> -4!<br> - 1

nexus9112 [7]

Answer:

$P(4,3) = 4!$

Step-by-step explanation:

Given

P(4,3)

Required

Solve

Using permutation formula;

$P(n,r) = \frac{n!}{(n-r)!}$

This implies that

$P(4,3) = \frac{4!}{(4-3)!}$

$P(4,3) = \frac{4!}{(1)!}$

$P(4,3) = \frac{4!}{1!}$

$P(4,3) = \frac{4!}{1}$

$P(4,3) = 4!$

6 0

3 years ago

Find the number to which the sequence {(3n+1)/(2n-1)} converges and prove that your answer is correct using the epsilon-N defini

Nat2105 [25]

By inspection, it's clear that the sequence must converge to $\dfrac32$ because

$\dfrac{3n+1}{2n-1}=\dfrac{3+\frac1n}{2-\frac1n}\approx\dfrac32$

when $n$ is arbitrarily large.

Now, for the limit as $n\to\infty$ to be equal to $\dfrac32$ is to say that for any $\varepsilon>0$ , there exists some $N$ such that whenever $n>N$ , it follows that

$\left|\dfrac{3n+1}{2n-1}-\dfrac32\right|$

From this inequality, we get

$\left|\dfrac{3n+1}{2n-1}-\dfrac32\right|=\left|\dfrac{(6n+2)-(6n-3)}{2(2n-1)}\right|=\dfrac52\dfrac1{|2n-1|}$
$\implies|2n-1|>\dfrac5{2\varepsilon}$
$\implies2n-1\dfrac5{2\varepsilon}$
$\implies n\dfrac12+\dfrac5{4\varepsilon}$

As we're considering $n\to\infty$ , we can omit the first inequality.

We can then see that choosing $N=\left\lceil\dfrac12+\dfrac5{4\varepsilon}\right\rceil$ will guarantee the condition for the limit to exist. We take the ceiling (least integer larger than the given bound) just so that $N\in\mathbb N$ .

6 0

3 years ago