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3 years ago

What does bidmass stant for in maths

Mathematics

2 answers:

aalyn [17]3 years ago

7 0

Bodmas= Bracket, of, division, multiplication, addition, subtraction

alex41 [277]3 years ago

3 0

Brackets, Indices, Division and Multiplication, Addition and Subtraction

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6.6÷2.2+(3.2+4)=<br> Use pemdas

maxonik [38]

3.2+4=7.2
6.6/2.2=3
3+7.2=10.2
10.2

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3 years ago

What is the product (Y squared plus 3Y +7)(AY squared plus Y +1)

Flauer [41]

Haha this one is the exactly why

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3 years ago

.78 3/11 as a percent

Anestetic [448]

.78 will be 78% and 3/11 will be 27% hope I helped!

8 0

2 years ago

PLEASEEEE HELP ME WITH THIS!!

dem82 [27]

Answer:

$x=\boxed {3.5}$

Step-by-step explanation:

Given:

AC = 5

BC = 6

DB = 4.2

∠ACD ≅ ∠BCD

AD = $x$

∵ ∠ACD ≅ ∠BCD

∴ CD is the angle Bisector of ∠ ACB

We know that, from angle bisector theorem that,

The angle bisector of an interior angle of a triangle divides the opposite side in two segments such that the two segments are proportional to the other two sides of the same triangle.

Therefore, the sides that are in proportion are given as:

$\dfrac{AC}{BC} =\dfrac{AD}{DB}$

Substituting the given values we get:

$\dfrac{5}{6} =\dfrac{x}{4.2}\\\\6x=5\times 4.2\\\\6x=21\\\\x=\dfrac{21}{6}=3.5\\\\\therefore x=3.5\ unit$

Therefore, the value of $x$ is 3.5.

4 0

3 years ago

Here are two expressions with a mixing number:

photoshop1234 [79]

An irrational number is a real number that can't be written as a fraction: since all rational numbers ( $\mathbb{Q}$ ) are real numbers ( $\mathbb{R}$ ), the irrational numbers are all elements in $\mathbb{R}$ that don't belong to $\mathbb{Q}$ .

The rational numbers are closed with respect to sum and multiplication, which means that the sum/product of two rational numbers is a rational number.

Note that, of all the options, only $\sqrt{3}$ is irrational, because the other options are

$3 = \dfrac{3}{1},\quad 2 =\dfrac{2}{1},\quad 0.\overline{23} = \dfrac{23}{99}$

so they all can be written as fraction. This means that

$-\dfrac{5}{9} \times 3,\quad -\dfrac{5}{9} \times 2,\quad -\dfrac{5}{9} \times\dfrac{23}{99}$

are all rational numbers, because they are the multiplication of two rational numbers.

So, your only hope to get an irrational number is to multiply

$-\dfrac{5}{9}\times \sqrt{3}$

Expression 2 works exactly in the same way, because everything we said about the product is true for the sum as well.

5 0

3 years ago