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3 years ago

PLZ HELP ME!!!!! I WILL MARK BRAINLIEST!! How is math related to language? Give two examples

Mathematics

2 answers:

tatyana61 [14]3 years ago

3 0

Answer:

alegbra and polynomial

navik [9.2K]3 years ago

3 0

English and mathematics as languages are similar and yet different. They both are languages of signs and symbols which combine to “words” “sentences” and “stories”. These “stories”have a grammar, which is universal and hence makes them languages of international communication.

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What are the exact values of the six trigonometric functions for -7pi/6 radians?

prohojiy [21]

For -7pi/6 is an angle in second quadrant, then sine and cosecant must be positive; and cosine, secant, tangent and cotangent must me negative.
The reference angle is:
7pi/6-pi=7pi/6-6pi/6=(7pi-6pi)/6=pi/6
Then
sin(-7pi/6)=sin(pi/6)→sin(-7pi/6)=1/2

cos(-7pi/6)=-cos(pi/6)→cos(-7pi/6)=-sqrt(3)/2

csc(-7pi/6)=1/sin(-7pi/6)=1/(1/2)=1(2/1)=2/1→csc(-7pi/6)=2

sec(-7pi/6)=1/cos(-7pi/6)=1/(-sqrt(3)/2)=-1(2/sqrt(3))=-2/sqrt(3)→
sec(-7pi/6)=-[2/sqrt(3)]*sqrt(3)/sqrt(3)=-2sqrt(3)/[sqrt(3)]^2→
sec(-7pi/6)=-2sqrt(3)/3

tan(-7pi/6)=sin(-7pi/6)/cos(-7pi/6)=(1/2)/(-sqrt(3)/2)=-(1/2)*(2/sqrt(3))→
tan(-7pi/6)=-2/[2sqrt(3)]=-1/sqrt(3)=-[1/sqrt(3)]*[sqrt(3)/sqrt(3)]→
tan(-7pi/6)=-sqrt(3)/[sqrt(3)]^2→tan(-7pi/6)=-sqrt(3)/3

cot(-7pi/6)=cos(-7pi/6)/sin(-7pi/6)=[-sqrt(3)/2]/(1/2)=-sqrt(3)/2*(2/1)→
cot(-7pi/6)=-2sqrt(3)/2→cot(-7pi/6)=-sqrt(3)

Answers:
sin(-7pi/6) = 1/2
cos(-7pi/6) = - sqrt(3)/2
tan(-7pi/6) = - sqrt(3)/3
csc(-7pi/6) = 2
sec(-7pi/6) = - 2*sqrt(3)/2
cot(-7pi/6) = - sqrt(3)

4 0

3 years ago

Read 2 more answers

The average fine for entering a national Forest wilderness without a permit increased from $65 to $78. What rate of increase is

andrezito [222]

20%

A change from 65 to 78 represents a positive change (increase) of 20%

Use the formula found below to find the percent change by replacing the given values:

Percent change = [(New - Old ) / |Old|] x 100%

3 0

3 years ago

Read 2 more answers

A rectangle has vertices W(2,3), X(2,6), Y(7,6), and Z(7,3).

KiRa [710]

Answer:

Step-by-step explanation:

A rectangle has 4 sides.

2 of them are lengths and 3 of them are widths.

We can simply use coordinate geometry (without graphing) to find side lengths of the rectangle. We will use Distance Formula.

We can find all the 4 lengths by using Distance Formula from points:

W and X

X and Y

Y and Z

W and Z

Note, that we don't need to find all 4 of them individually, because 2 are lengths (same) and 2 are widths (same). Thus we can find

Distance of WX, which would be same as distance of YZ

also

Distance of XY which would be same as distance of WZ

<em><u>Note:</u></em> Distance Formula is $D=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}$ where D is the distance, x_1, y_1 is the first coordinate points and x_2,y_2 is the second coordinate points

4 0

4 years ago

there are 1012 souvenir paperweights that need to be packed in boxes each box will hold 12 paperweights how many boxes will be n

Rzqust [24]

1012 divided by 120 is 8.43 so 9 boxes will be needed

5 0

4 years ago

Prove that the diagonals of a parallelogram bisect each other

Nady [450]

Answer:

[ See the attached picture ]

The diagonals of a parallelogram bisect each other.

✧ Given : ABCD is a parallelogram. Diagonals AC and BD intersect at O.

✺ To prove : AC and BD bisect each other at O , i.e AO = OC and BO = OD.

Proof : $\begin{array}{ |c| c | c | } \hline \tt{SN}& \tt{STATEMENTS} & \tt{REASONS}\\ \hline 1& \sf{In \: \triangle ^{s} \:AOB \: and \: COD } \\ \sf{(i)}& \sf{ \angle \: OAB = \angle \: OCD\: (A)}& \sf{AB \parallel \: DC \: and \: alternate \: angles} \\ \sf{(ii)} &\sf{AB = DC(S)}& \sf{Opposite \: sides \: of \: a \: parallelogram} \\ \sf{(iii)} &\sf{ \angle \: OBA= \angle \: ODC(A)} &\sf{AB \parallel \:DC \: and \: alternate \: angles} \\ \sf{(iv)}& \sf{ \triangle \:AOB\cong \triangle \: COD}& \sf{A.S.A \: axiom}\\ \hline 2.& \sf{AO = OC \: and \: BO = OD}& \sf{Corresponding \: sides \: of \: congruent \: triangle}\\ \hline 3.& \sf{AC \: and \: BD \: bisect \: each \: other \: at \: O}& \sf{From \: statement \: (2)}\\ \\ \hline\end{array}. Proved ✔$

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5 0

3 years ago