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4 years ago

An angle that has a measure greater than 0⁰ and less than 90⁰

Mathematics

2 answers:

Vitek1552 [10]4 years ago

8 0

Answer:

Obtuse

Step-by-step explanation:

The obtuse has a over 90 ° angle and is greater than that

prohojiy [21]4 years ago

6 0

Answer:

Acute angle

Step-by-step explanation:

An angle that has a measure greater than 0⁰ and less than 90 is known as acute angle

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State whether each percent of change is a percent increase or a percent decrease. Then find the percent of increase or decrease

Otrada [13]

Percent decrease. 41%
percent change= (new-old)/old *100
= -41 % the negative means decreasing

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3 years ago

Increase £3.20 in the ratio 2:3<br>Please show your working out (/formulae).

Degger [83]

Answer:

£4.80

Step-by-step explanation:

Total amount = £3.20

Ratio = 2 : 3

Old : New = 2 : 3

3.20 : x = 2:3

$\frac{3.20}{x}=\frac{2}{3}\\\\3.20*3=2*x\\\\\frac{3.20*3}{2}=x\\\\x = 1.60 *3\\$

x = £ 4.80

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3 years ago

What is the decomposition of 43.586

Fiesta28 [93]

43.586- anything 5 and more rounds up, so the 6 rounds the 8 up to 9, the 9 rounds the 5 up to a 6, so your answer wound be 43.6 because 43.5(tenths)8(hundredths)6(thousandths), i hope this helps

Leave a thanks :D

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3 years ago

In an experiment a six-sided die is rolled a number of times. The results are shown below.

Sergeu [11.5K]

The answer is D hope this helps!!

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3 years ago

F(x)=1/2×+6 plz helpppppppppp

FromTheMoon [43]

Answer:

The solution set can be given as:

$\{x|x\ \epsilon\ R,x\neq -3\}$

Step-by-step explanation:

Given function:

$f(x)=\frac{1}{2x+6}$

To find the domain of the function in set notation.

Solution:

For the function $f(x)$ to exist the denominator must be ≠ 0

We have the denominator $(2x+6)$ which cannot be = 0.

Thus, we can find the domain of the function using the above relation.

The function $f(x)$ will not exist when:

$2x+6=0$

Solving for $x$

Subtracting both sides by 6.

$2x+6-6=0-6$

$2x=-6$

Dividing both sides by 2.

$\frac{2x}{2}=\frac{-6}{2}$

∴ $x=-3$

Thus, the function will not exist at $x=-3$ . This means it has all real number solutions except -3.

The solution set can be given as:

$\{x|x\ \epsilon\ R,x\neq -3\}$

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4 years ago