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ludmilkaskok [199]

3 years ago

6

Difference between an integer and a real number

2 answers:

notsponge [240]3 years ago

4 0

Guguvyftcgxfctc ull busy

jenyasd209 [6]3 years ago

3 0

Answer:

Real number is any number. This includes integers, rational, and irrational numbers.

While integers are just positive or negative whole numbers that also includes 0.

Hope this helps!

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Change the following mixed expressions to fractions and then simplify the complex fractions.

Zielflug [23.3K]

Answer:

X=2

Step-by-step explanation:

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3 years ago

A winch is being used to pull a 3500-lb vehicle 30'. How much work is performed?

AlekseyPX

Answer:

105,000 foot-pounds.

Step-by-step explanation:

Work done = 3500 * 30 foot pounds

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3 years ago

Find X. please i need help with this

-BARSIC- [3]

Answer:

Step-by-step explanation:

X + X + 96 = 180 {SUM OF ANGLES OF TRIANGLE}

2X + 96 = 180

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4 0

3 years ago

Read 2 more answers

I don’t know how to explain on why the student was incorrect

Zolol [24]

Answer:

the constant proportionality means the ratio between two directly proportional quantities.

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7 0

3 years ago

Directions: Find the indicated trigonometric ratio as a fraction in simplest form. 1. Sin L =

Rainbow [258]

Answer:

$\sin L = 0.60$

$tan\ N = 1.33$

$\cos L = 0.80$

$\sin N = 0.80$

Step-by-step explanation:

Given

See attachment

From the attachment, we have:

$MN = 6$

$LN = 10$

First, we need to calculate length LM,

Using Pythagoras theorem:

$LN^2 = MN^2 + LM^2$

$10^2 = 6^2 + LM^2$

$100 = 36 + LM^2$

Collect Like Terms

$LM^2 = 100 - 36$

$LM^2 = 64$

$LM = 8$

Solving (a): $\sin L$

$\sin L = \frac{Opposite}{Hypotenuse}$

$\sin L = \frac{MN}{LN}$

Substitute values for MN and LN

$\sin L = \frac{6}{10}$

$\sin L = 0.60$

Solving (b): $tan\ N$

$tan\ N = \frac{Opposite}{Adjacent}$

$tan\ N = \frac{LM}{MN}$

Substitute values for LM and MN

$tan\ N = \frac{8}{6}$

$tan\ N = 1.33$

Solving (c): $\cos L$

$\cos L = \frac{Adjacent}{Hypotenuse}$

$\cos L = \frac{LM}{LN}$

Substitute values for LN and LM

$\cos L = \frac{8}{10}$

$\cos L = 0.80$

Solving (d): $\sin N$

$\sin N = \frac{Opposite}{Hypotenuse}$

$\sin N = \frac{LM}{LN}$

Substitute values for LM and LN

$\sin N = \frac{8}{10}$

$\sin N = 0.80$

7 0

3 years ago