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1 year ago

PQRS is a quadilateral in which <PQR=74°;

Mathematics

1 answer:

8090 [49]1 year ago

8 0

Hey bro not the best but remember too use PEMDAS

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in the previous question, what is the distance from a side of the triangle to the incenter? can someone also explain how to get

denis23 [38]

In centre is the centre of a circle that is inscribed in a triangle, called the incircle, i.e. circle is tangent to all three sides AND located inside the triangle. An incircle is roughly sketched in the accompanying diagram.

Since all radii of an incircle are equal, we have
4x-1=6x-5
Solving for x,
6x-4x=5-1
x=2
consequently, distance from incentre to each side of the triangle is
4(2)-1=7
check 6x-5=6(2)-5=7... checks.

Answers:
The value of x = 2
The distance from incentre to side of triangle is 7

5 0

3 years ago

Solve this equation. Enter your answer in the box.<br> 13(y+7)=3(y−1).

Fofino [41]

★To Solve :

$\rm\blue{13(y+7) = 3(y-1)}$

⠀⠀⠀⠀⠀

★SOLUTION :

$\rm\red\leadsto{13(y+7) = 3(y-1)}$

$\rm\red\leadsto{13\times{y}+13\times7= 3\times{y}-3\times1}$

$\rm\red\leadsto{13y+91=3y-3}$

$\rm\red\leadsto{13y-3y=-3-91}$

$\rm\red\leadsto{10y=-94}$

$\rm\red\leadsto{} y =\dfrac{ { \cancel{-94}}^{ \:-45 } }{ { \cancel{10}}^{ \: 5} }$

$\rm\red\leadsto{y=}\dfrac{-47}{5}$

____________________

5 0

3 years ago

Read 2 more answers

How to find the base of an isosceles when given one side

MA_775_DIABLO [31]

Double the side and subtract by 360

3 0

2 years ago

1( below is the pre image dilate by a factor of 2

bonufazy [111]

Before:It's too short. Write at least 20 characters to explain it well.

After:It's too short. Write at least 20 characters to explain it well.

3 0

2 years ago

Suppose the graph of a cubic polynomial function has the same zeroes and passes through the coordinate (0, –5). Describe the ste

olga_2 [115]

1. "the graph has the same zeros" : so let a be the "triple" root of the cubic polynomial function.

2. So f(x)= $(x-a)^{3}$

3. Don't forget that the expression might have a coefficient b as well, and still maintain the conditions:

$f(x)=b(x-a)^{3}$

4. Now, f(0)=-5 so $-5=f(0)=b(0-a)^{3}=b(-a) ^{3}=-ba ^{3}$

$-5=-ba ^{3}$

$5=ba ^{3}$

$b= \frac{5}{ a^{3} }$

5. the function is $f(x)=\frac{5}{ a^{3} }(x-a)^{3}$ where a can be any real number except 0

7 0

3 years ago

Read 2 more answers