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3 years ago

Solve the simultaneous equations y = 2x^2 y = 3x + 14

Mathematics

1 answer:

Mekhanik [1.2K]3 years ago

3 0

Answer:

x = -2

And, x = 7 ÷ 2

Step-by-step explanation:

Given that

y = 2x^2

y = 3x + 14

Now equate these two above equations

2x^2 = 3x + 14

2x^2 - 3x - 14

2x^2 -7x + 4x - 14

x(2x - 7) + 2(2x - 7)

(x + 2) (2x - 7)

So it can be x = -2

And, x = 7 ÷ 2

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3 years ago

A parking garage charges $5.50 to park for 4 hours and $7.75 to park for 7 hours. If the cost is a linear function of the number

Veseljchak [2.6K]

Given:
hours parking charge
4 5.50
7 7.75

7 - 4 = 3 hours
7.75 - 5.50 = 2.25

2.25 / 3 = 0.75

0.75 is the change in amount for every hour spent. It is the variable unit rate

4 hours is 5.50.

Cost of parking for 3 hours is: 5.50 - 0.75 = 4.75

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3 years ago

Read 2 more answers

Directions: Find the Surface Area

kenny6666 [7]

Answer:

Surface area = 92 ft²

Step-by-step explanation:

Formula to find the surface area of a rectangular prism is,

Surface area = 2(lb + bh + hl)

Here, l = length of the prism

b = width of the prism

h = height of the prism

By substituting the measures of the sides given in the picture,

Surface area = 2(2×3 + 3×8 + 8×2)

= 2(6 + 24 + 16)

= 2(46)

= 92 square feet

Therefore, surface area of the given rectangular prism is 92 ft².

4 0

3 years ago

Please show work and thank youuu

Papessa [141]

Answer:

x = 8√2

Step-by-step explanation:

As the opposing side of the angle and the hypotenuse are given, take the sine ratio of the angle.

sin 45° = x/16
1/√2 = x/16
x = 16 / √2
x = 16√2 / 2
x = 8√2

8 0

2 years ago

Read 2 more answers

Prove that: (b²-c²/a)CosA+(c²-a²/b)CosB+(a²-b²/c)CosC = 0

IRISSAK [1]

Prove that:

$\:\:\sf\:\:\left(\dfrac{b^2-c^2}{a}\right)\cos A+\left(\dfrac{c^2-a^2}{b}\right)\cos B +\left(\dfrac{a^2-b^2}{c}\right)\cos C=0$

Proof: 

We know that, by Law of Cosines,

$\sf \cos A=\dfrac{b^2+c^2-a^2}{2bc}$
$\sf \cos B=\dfrac{c^2+a^2-b^2}{2ca}$
$\sf \cos C=\dfrac{a^2+b^2-c^2}{2ab}$

Taking LHS

$\left(\dfrac{b^2-c^2}{a}\right)\cos A+\left(\dfrac{c^2-a^2}{b}\right)\cos B +\left(\dfrac{a^2-b^2}{c}\right)\cos C$

Substituting the value of cos A, cos B and cos C,

$\longmapsto\left(\dfrac{b^2-c^2}{a}\right)\left(\dfrac{b^2+c^2-a^2}{2bc}\right)+\left(\dfrac{c^2-a^2}{b}\right)\left(\dfrac{c^2+a^2-b^2}{2ca}\right)+\left(\dfrac{a^2-b^2}{c}\right)\left(\dfrac{a^2+b^2-c^2}{2ab}\right)$

$\longmapsto\left(\dfrac{(b^2-c^2)(b^2+c^2-a^2)}{2abc}\right)+\left(\dfrac{(c^2-a^2)(c^2+a^2-b^2)}{2abc}\right)+\left(\dfrac{(a^2-b^2)(a^2+b^2-c^2)}{2abc}\right)$

$\longmapsto\left(\dfrac{(b^2-c^2)(b^2+c^2)-(b^2-c^2)(a^2)}{2abc}\right)+\left(\dfrac{(c^2-a^2)(c^2+a^2)-(c^2-a^2)(b^2)}{2abc}\right)+\left(\dfrac{(a^2-b^2)(a^2+b^2)-(a^2-b^2)(c^2)}{2abc}\right)$

$\longmapsto\left(\dfrac{(b^4-c^4)-(a^2b^2-a^2c^2)}{2abc}\right)+\left(\dfrac{(c^4-a^4)-(b^2c^2-a^2b^2)}{2abc}\right)+\left(\dfrac{(a^4-b^4)-(a^2c^2-b^2c^2)}{2abc}\right)$

$\longmapsto\dfrac{b^4-c^4-a^2b^2+a^2c^2}{2abc}+\dfrac{c^4-a^4-b^2c^2+a^2b^2}{2abc}+\dfrac{a^4-b^4-a^2c^2+b^2c^2}{2abc}$

On combining the fractions,

$\longmapsto\dfrac{(b^4-c^4-a^2b^2+a^2c^2)+(c^4-a^4-b^2c^2+a^2b^2)+(a^4-b^4-a^2c^2+b^2c^2)}{2abc}$

$\longmapsto\dfrac{b^4-c^4-a^2b^2+a^2c^2+c^4-a^4-b^2c^2+a^2b^2+a^4-b^4-a^2c^2+b^2c^2}{2abc}$

Regrouping the terms,

$\longmapsto\dfrac{(a^4-a^4)+(b^4-b^4)+(c^4-c^4)+(a^2b^2-a^2b^2)+(b^2c^2-b^2c^2)+(a^2c^2-a^2c^2)}{2abc}$

$\longmapsto\dfrac{(0)+(0)+(0)+(0)+(0)+(0)}{2abc}$

$\longmapsto\dfrac{0}{2abc}$

$\longmapsto\bf 0=RHS$

LHS = RHS proved.

7 0

3 years ago