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Leya [2.2K]
2 years ago
13

What is the solution(s) to the system of equations y = x² +4 and y = 2x + 4

Mathematics
1 answer:
bija089 [108]2 years ago
3 0

Answer:

(x,y) = (0,4)~ \text{and}~ (x,y) = (2,8)

Step by step explanation:

y = x^2 +4~~~~~~~~~...(i)\\\\y = 2x +4~~~~~~~~~...(ii)\\\\\text{From (i) and (ii):}\\\\~~~~~~x^2 +4 = 2x+4\\\\\implies x^2 = 2x\\\\\implies x^2 -2x = 0\\\\\implies x(x-2) = 0\\\\\implies x =0,~ x = 2

\text{Substitute x = 0 in eq (i):}\\\\y = 0^2 +4 = 4\\\\\text{Substitute x = 2 in eq (i):}\\\\y=2^2+4 = 4+4 = 8\\\\\text{Hence,}~ (x,y)= \{(2,8), (0,4)\}

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Answer:

ok i ned time

Step-by-step explanation:

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2 years ago
-10(10x-12)=-9(-9x-2)-5
Viefleur [7K]

10(10x−12)=−9(−9x−2)−5
Step 1: Simplify both sides of the equation.
−10(10x−12)=−9(−9x−2)−5
(−10)(10x)+(−10)(−12)=(−9)(−9x)+(−9)(−2)+−5(Distribute)
−100x+120=81x+18+−5
−100x+120=(81x)+(18+−5)(Combine Like Terms)
−100x+120=81x+13
−100x+120=81x+13
Step 2: Subtract 81x from both sides.
−100x+120−81x=81x+13−81x
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Step 3: Subtract 120 from both sides.
−181x+120−120=13−120
−181x=−107
Step 4: Divide both sides by -181.
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−181
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Answer:
x=
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181

7 0
4 years ago
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Find the equation of the line
artcher [175]

Answer:

y=3x+3

Step-by-step explanation:

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3 years ago
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Find the dimensions of the open rectangular box of maximum volume that can be made from a sheet of cardboard 21 in. by 12 in. by
a_sh-v [17]

Answer:

Dimension of the box is 16.1\times 7.1\times 2.45

The volume of the box is 280.05 in³.

Step-by-step explanation:          

Given : The open rectangular box of maximum volume that can be made from a sheet of cardboard 21 in. by 12 in. by cutting congruent squares from the corners and folding up the sides.

To find : The dimensions and the volume of the box?

Solution :

Let h be the height of the box which is the side length of a corner square.

According to question,

A sheet of cardboard 21 in. by 12 in. by cutting congruent squares from the corners and folding up the sides.

The length of the box is L=21-2h

The width of the box is W=12-2h

The volume of the box is V=L\times W\times H

V=(21-2h)\times (12-2h)\times h

V=(21-2h)\times (12h-2h^2)

V=252h-42h^2-24h^2+4h^3

V=4h^3-66h^2+252h

To maximize the volume we find derivative of volume and put it to zero.

V'=12h^2-132h+252

0=12h^2-132h+252

Solving by quadratic formula,

h=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

h=\frac{-(-132)\pm\sqrt{132^2-4(12)(252)}}{2(12)}

h=\frac{132\pm72.99}{24}

h=2.45,8.54

Now, substitute the value of h in the volume,

V=4h^3-66h^2+252h

When, h=2.45

V=4(2.45)^3-66(2.45)^2+252(2.45)

V\approx 280.05

When, h=8.54

V=4(8.54)^3-66(8.54)^2+252(8.54)

V\approx -170.06

Rejecting the negative volume as it is not possible.

Therefore, The volume of the box is 280.05 in³.

The dimension of the box is

The height of the box is h=2.45

The length of the box is L=21-2(2.45)=16.1

The width of the box is W=12-2(2.45)=7.1

So, Dimension of the box is 16.1\times 7.1\times 2.45

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3 years ago
What are the side lengths of a 45-45-90 triangle with a hypotenuse of 10root2?
scoundrel [369]

Answer:

10

Step-by-step explanation:

let the lengths of sides=x (each)

x²+x²=(10√2)²

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x²=100

x=√100=10

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