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Genrish500 [490]
4 years ago
5

How is the location of a point on a grid described?

Mathematics
1 answer:
Andrews [41]4 years ago
5 0
So,

A point on a Cartesian Coordinate System is normally called an ordered pair.  It is normally written in the format (x,y).  For example the point/ordered pair (7,-8) means to move 7 units to the right of zero, and then 8 units down.
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What for omit mean in math terms?
LiRa [457]
Omit means: "to leave out" or "not include"
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4 years ago
Read 2 more answers
What is the value of x
Crank

Answer:

Step-by-step explanation:

Looking at the diagram,

Line DT is parallel to line line BM, therefore, triangle DHT is similar to triangle BHM. This means that

DH/DB = TH/MH

DH = 15 + 6 = 21

TH = 9 + x

Therefore

(9 + x)/21 = 9/6

Cross multiplying, it becomes

6(9 + x) = 21 × 9 = 189

54 + 6x = 189

Subtracting 54 from the left hand side and the right hand side of the equation, it becomes

6x + 54 - 54 = 189 - 54

6x = 135

Dividing the left hand side and the right hand side of the equation by 6, it becomes

6x/6 = 135/6

x = 22.5

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4 years ago
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What type of correlation (association)? The outside temperature and the amount of layers you wear.​
kykrilka [37]

Answer:

negitive corralation

Step-by-step explanation:

7 0
3 years ago
the Marked price of a book is rupees 160 if it is sold at 15% discount find the discount amount and selling price​
Ostrovityanka [42]

answer is 15%

Step-by-step explanation:

cost prices of book =160

selling prices of book =136

now,rate of discount =loss% (as discount only leads to some kind of loss)

so CP>SP

so loss=CP-SP =160-136=24

loss%= (loss/CP)÷100

=(24/160)÷100

=15%

so rate of discount =15%

7 0
3 years ago
In traveling across flat land, you see a mountain directly in front of you. Its angle of elevation (to the peak) is 3.5°. After
Harlamova29_29 [7]
Check the picture below.

\bf tan(3.5^o)=\cfrac{y}{15+x}\implies 15+x=\cfrac{y}{tan(3.5^o)}\\\\\\ \boxed{x=\cfrac{y}{tan(3.5^o)}-15}
\\\\\\
tan(9^o)=\cfrac{y}{x}\implies x\cdot  tan(9^o)=y\\\\\\ \left[ \cfrac{y}{tan(3.5^o)}-15 \right]tan(9^o)=y
\\\\\\
\left[ \cfrac{y-15tan(3.5^o)}{tan(3.5^o)} \right]tan(9^o)=y
\\\\\\

\bf \cfrac{y\cdot  tan(9^o)-15tan(3.5^o)tan(9^o)}{tan(3.5^o)}=y
\\\\\\
y\cdot  tan(9^o)-15tan(3.5^o)tan(9^o)=y\cdot tan(3.5^o)
\\\\\\
ytan(9^o)-ytan(3.5^o)=15tan(3.5^o)tan(9^o)
\\\\\\
\stackrel{common~factor}{y}[tan(9^o)-tan(3.5^o)]=15tan(3.5^o)tan(9^o)
\\\\\\
y=\cfrac{15tan(3.5^o)tan(9^o)}{tan(9^o)-tan(3.5^o)}

make sure your calculator is in Degree mode.

the amount is small, but bear in mind that is mile units.

3 0
3 years ago
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