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AysviL [449]
3 years ago
7

X=y x+2y =3 Solve using substitution

Mathematics
1 answer:
Zielflug [23.3K]3 years ago
3 0

Answer:

Therefore the Final Solution is

x = 1 and y = 1

Step-by-step explanation:

Given:

x=y                ...............Equation ( 1 )

x+2y=3         ................Equation ( 2 )

To Find:

x = ?

y = ?

Solution:

Equating Equation  ( 1 ) in Equation ( 2 ) we get

y+2y=3\\\\3y=3\\\\y=\frac{3}{3} \\\\\therefore y =1

Now Substituting y = 1 in Equation ( 1 ) we get

x =1\\\\\therefore x=1

Therefore the Final Solution is

x = 1 and y = 1

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What does 5 mean? What is the value of 5
kap26 [50]

Answer:

the absolute value of 5 is simply, 5

Step-by-step explanation:

6 0
3 years ago
Equilateral triangle ABC has an area of \sqrt{3}√ ​3 ​ ​​ . If the shaded region has an area of \piπK − \sqrt{3}√ ​3 ​ ​​ , what
Liono4ka [1.6K]

Answer:

The value of k = 4/3

Step-by-step explanation:

* Lets explain how to solve the problem

- An equilateral triangle ABC is inscribed in a circle N

- The area of the triangle is √3

- The shaded area is the difference between the area of the circle

  and the area of the equilateral triangle ABC

- The shaded are = k π - √3

- We need to find the value of k

* <u><em>At first lets find the length of the side of the Δ ABC</em></u>

∵ Δ ABC is an equilateral triangle

∴ Its area = √3/4 s² , where s is the length of its sides

∵ The area of the triangle = √3

∴ √3/4 s² = √3

- divide both sides by √3

∴ 1/4 s² = 1

- Multiply both sides by 4

∴ s² = 4 ⇒ take √ for both sides

∴ s = 2

∴ The length of the side of the equilateral triangle is 2

* <u><em>Now lets find the radius of the circle</em></u>

- In the triangle whose vertices are A , B and N the center of the circle

∵ AN and BN are radii

∴ AN = BN = r , where r is the radius of the circle

∵ The sides of the equilateral angles divides the circle into 3 equal

   arcs in measure where each arc has measure 360°/3 = 120°

∵ The measure of the central angle in a circle equal the measure

  of the its subtended arc arc

∵ ∠ANB is an central angle subtended by arc AB

∵ The measure of arc AB is 120°

∴ m∠ANB = 120°

- By using the cosine rule in Δ ANB

∵ AB = 2 , AN = BN = r , m∠ANB = 120°

∴ (2)^{2}=r^{2}+r^{2}-2(r)(r)cos(120)

∴ 4=r^{2}+r^{2}-2(r)(r)(-0.5)

∴ 4=r^{2}+r^{2}-(-r^{2})

∴ 4=r^{2}+r^{2}+r^{2}

∴ 4=3r^{2}

- Divide both sides by 3

∴ r^{2}=\frac{4}{3}

- Take square root for both sides

∴ r = 2/√3

* <u><em>Lets find the value of k</em></u>

∵ Area circle = πr²

∵ r = 2/√3

∴ Area circle = π(2/√3)² = (4/3)π

∵ Area shaded = area circle - area triangle

∵ Area triangle = √3

∴ Area shaded = (4/3) π - √3

∵ Area of the shaded part is π k - √3

- Equate the two expressions

∴ π k - √3 = (4/3) π - √3

∴ k = 4/3

* The value of k = 4/3

7 0
3 years ago
Help me plzzz i don't know what to put​
Pachacha [2.7K]
The correct answer should be 4. 2 times 0.3 is 0.6. 3.4 plus 0.6 equals 4. ;)
7 0
3 years ago
Find the value of c that makes t^2– 22t+ c a perfect square trinomial.
riadik2000 [5.3K]

Answer:

ZkgJHhHJ HHha sorry i am not sure

3 0
3 years ago
Consider the vectors a =3i +j −k, b =i +j +4k, c=i +3j +k, d =−i −3j +k, g =−3i −j +k. Which pairs (if any) of these vectors are
11111nata11111 [884]

Answer:

a and b are perpendicular to each other, as are b and d, b and g

Step-by-step explanation:

To check whether two vectors are perpendicular to each other, we need the angle between these vectors to be 90 degrees.

We can find the angle between to vectors a and b from the following relation:

The cosine of the angle \theta between two vectors is equal to the dot product of this vectors divided by the product of vector magnitude.

So

cos(\theta) = \frac{a.b}{|a||b|}

cos(90) = 0, so when the dot product between vectors a and b is 0, it means that these vectors are perpendicular to each other.

Now, for your exercise, let's compute the dot product between these vectors.

-----------

a.b = (3,1,-1).(1,1,4) = 3+1-4 = 0

So a and b are perpendicular to each other.

------------

a.c = (3,1,-1).(1,3,1) = 3+3-1 = 5

a and c are not perpendicular to each other.

--------------

a.d = (3,1,-1).(-1,-3,1) = -3-3-1 = -7

So not perpendicular

---------------

a.g = (3,1,-1).(-3,-1,1) = -9-1-1 = -11

Not

-----------------

b.c = (1,1,4).(1,3,1) = 1+3+4 = 8

Not

------------------

b.d = (1,1,4).(-1,-3,1) = -1 -3 +4 = 0

b.d = 0, so b and d are perpendicular to each other

--------------------

b.g = (1,1,4).(-3,-1,1) = -3-1+4 = 0

b.g = 0, perpendicular

---------------------

c.d = (1,3,1).(-1,-3,1) = -1-9+1 = -9

No

-----------------------

c.g = (1,3,1).(-3,-1,1) = -3-3+1 = -5

No

------------------------

d.g = (-1,-3,1).(-3,-1,1) = 3+3+1 = 7

No

7 0
3 years ago
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