All categories

2 years ago

Given the linear equation r=-2.5 t and t can only be a negative number, what quandrant will the r coordinate be located?

Mathematics

2 answers:

Citrus2011 [14]2 years ago

4 0

Answer:

quadrant 4 i think

Step-by-step explanation:

sergiy2304 [10]2 years ago

3 0

Answer:

Actually the correct answer is 2

Step-by-step explanation:

You might be interested in

6/3 + (-1/6) = ? (These are supposed to be fractions)

Phoenix [80]

Answer: 11/6 or 1 5/6

Simplify 6/3

6/3÷3/3=2/1

Change 2/1

2/1×6/6=12/6

New Problem: 12/6+(-1/6)

Add

12+-1/6=11/6

Simplify

11/6 = 1 5/6

6 0

2 years ago

In the given figure ABCD is a tripepizeum with AB||CD. If AO = x-1, CO = BO = x+1 and CD = x+4. Find the value of x.

Hitman42 [59]

$\longmapsto$ The value of "x" is 5.

$\large\underline{\sf{Solution-}}$

Given that,

A trapezium ABCD in which AB || CD such that

AO = x - 1

CO = BO = x + 1

OD = x + 4

Now,

$\rm In \: \triangle \: AOB \: and \: \triangle \: COD$

$\rm \: \angle \: AOB \: and \: \angle \: COD \: \: \{vertically \: opposite \: angles \}$

$\rm \: \angle \:ABO \: and \: \angle \: CDO \: \: \{alternate \: interior \: angles \}$

$\bf \: \triangle \: AOB \: \sim \: \triangle \: COD \: \: \: \{AA \: similarity \}$

$\bf\longmapsto\:\dfrac{AO}{CO} = \dfrac{BO}{DO}$

$\rm \longmapsto\:\dfrac{x - 1}{x + 1} = \dfrac{x + 1}{x + 4}$

$\rm \longmapsto\:(x - 1)(x + 4) = {(x + 1)}^{2}$

$\rm \longmapsto\: {x}^{2} - x + 4x - 4 = {x}^{2} + 1 + 2x$

$\rm \longmapsto\: 3x - 4 = 1 + 2x$

$\rm \longmapsto\: 3x - 2x= 1 +4$

$\bf\longmapsto \:x = 5$

7 0

2 years ago

The teenagers of a school sold 430 tickets for their Christmas program. The girls sold 15% more tickets than the boys sold. How

Paul [167]

The girls sold 230
The boys sold 200

8 0

3 years ago

What are the next 6 terms 1,3,9,27 ?

Triss [41]

Next 6 terms =81 ,243,729,2187,6561,19683

4 0

3 years ago

Find the volume of a trough 5 meters long whose ends are equilateral triangles, each of whose

yawa3891 [41]

Answer:

$V=5\sqrt{3}\ m^3$

Step-by-step explanation:

we know that

The volume of a trough is equal to

$V=BL$

where

B is the area of equilateral triangle

L is the length of a trough

step 1

Find the area of equilateral triangle B

The area of a equilateral triangle applying the law of sines is equal to

$B=\frac{1}{2} b^{2} sin(60\°)$

where

$b=2\ m$

$sin(60\°)=\frac{\sqrt{3}}{2}$

substitute

$B=\frac{1}{2}(2)^{2} (\frac{\sqrt{3}}{2})$

$B=\sqrt{3}\ m^{2}$

step 2

Find the volume of a trough

$V=BL$

we have

$B=\sqrt{3}\ m^{2}$

$L=5\ m$

substitute

$V=(\sqrt{3})(5)$

$V=5\sqrt{3}\ m^3$

3 0

3 years ago