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2 years ago

Can someone please help me?

Mathematics

1 answer:

Andru [333]2 years ago

4 0

Answer:

Use the SINE COSINE and the TANGENT rule

Step-by-step explanation:

1.1st use the Pythagoras theorem to find the value of the other side

2.Now be comfortable to use the rule............................

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Find the solution to the set <br> b^2+12b=0 <br> use a comma

shepuryov [24]

Factor
b(b+12)=0
set to zero
b=0

b+12=0
b=-12

b=0 or -12
written as this way
b=-12,0

5 0

3 years ago

HLEPPPPPP!!!!!

Ksenya-84 [330]

Answer:

Solutions:-5,0,9,6

Not solutions: 12.5,10

Explanation:

The inequality that was given is:

20 > 5 + 3x

First, group similar terms;

20 - 5 < 3x

15 < 3x

x = 5

Thus, the solutions are:-3.5, 5, 4 3/4, -15

The not solutions are:-10 2/3, 129,

Hope this helps! :)

7 0

3 years ago

Helpppppppppppp ppppp:/

jeka94

Answer:

Step-by-step explanation:

I hope this helps

3 0

3 years ago

Give me an example of an irrational number that is greater than 10

Pavel [41]

To find:

An irrational number that is greater than 10.

Solution:

Irritation number: It cannot be expression in the form of $\dfrac{p}{q}$ , where, $q\neq 0, p,q$ are integers.

For example: $\sqrt{2}\sqrt{3},\pi, 1.263689...,etc.$ .

We know that square of 10 is 100. So, square root of any prime number is an example of an irrational number that is greater than 10.

First prime number after 100 is 101.

Required irrational number $=\sqrt{101}$

Therefore, $\sqrt{101}$ is an irrational number that is greater than 10.

5 0

3 years ago

Is (-2,3/4,-2/7,0) a set of real numbers

DedPeter [7]

Answer:

Yes

Step-by-step explanation:

All of these numbers are real numbers. None of them are complex numbers or irrational like <em>i</em>, s the answer is yes.

I have provided a chart to help you out with this topic.

Hope this helps! :)

8 0

3 years ago