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3 years ago

WILL MARK BRAINLIEST ANSWER 2x^2+20x=-38 show work

2 answers:

8 0

2x^2+20x = -38

2x^2 + 20x + 38 = 0

Divide both sides by 2.

x^2 + 10x + 19 = 0

The quadratic formula works very well here.

Let a = 1, b = 10 and c = 19

x = [-b +- sqrt{b^2 - 4ac}]/(2a)

Plug those values into the formula to find two answers for x.

One is negative and the other positive.

Good luck.

tatuchka [14]3 years ago

5 0

Move 38 to the left side of the equation by adding it to both sides.
2x ^2 + 20x + 38 = 0
Use the quadratic formula to find the solutions.
−b±√b^2 − 4 (ac)/2a
Substitute a = 2, b = 20, and c = 38 into the quadratic formula and solve for x.
−20±√20 ^2 − 4⋅(2⋅38)/2⋅2
Simplify. x = −5±√6
The result can be shown in both exact and approximate form.
x = −5±√6
x ≈ −2.55051025, −7.44948974

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What is the sum of 3 consecutive integers of 87

Leviafan [203]

Hey there! Hello!

So, to set this problem up, all you need to do is imagine that the smallest number of these consecutive integers is x. The next two numbers will be x+1, then x+2, and that, altogether, will equal 87. In a problem, this will look like this:

$x+x+1+x+2=87$

All we need to do is solve for x! We can simplify the problem by combining like terms:

$x+x+x+1+2=3x+3=87$
$3x+3-3=87-3$
$3x/3=84/3$
$x=28$

Now that we know what x is, we can plug it into our original problem:

$x+x+1+x+2 = 28+29+30=87$

Hope this helped you out! Feel free to ask me any additional questions if you need further clarification. :-)

3 0

3 years ago

Functions f(x) and g(x) are composed to form h (x) = startroot x cubed minus 2 endroot. if f (x) = startroot x 2 endroot and g (

iren [92.7K]

Function Composition exists when two functions f(x) and g(x) result in another function h(x), such that h(x) = f(g(x)). For function composition h(x) = f(g(x)), the value of a exist -4.

<h3>What is function Composition?</h3>

Function Composition exists when two functions f(x) and g(x) result in another function h(x), such that h(x) = f(g(x)). In other words, put the outcome of one function into the other one.

Here, h(x) = f(g(x))

which means, h(x) = f(x³+ a)

$$h(x) = \sqrt{x^3+a+2}$

$$\sqrt{x^3-2} = \sqrt{x^3+a+2}$

To estimate the value of a, we separate equal terms:

1) Both are squared, so we can "eliminate" the square;

2) x³ = x³

3) -2 = a+2

a = -4

For function composition h(x) = f(g(x)), the value of a exist -4.

To learn more about function composition refer to:

brainly.com/question/17299449

#SPJ4

8 0

2 years ago

For a triangle $XYZ$, we use $[XYZ]$ to denote its area. Let $ABCD$ be a square with side length $1$. Points $E$ and $F$ lie on

nata0808 [166]

An algebraic equation enables the expression of equality between variable expressions

$\underline{The \ value \ of \ [AEF] \ is \ \dfrac{4}{9}}$

The reason the above value is correct is given as follows:

The given parameters are;

The symbol for the area of a triangle ΔXYZ = [XYZ]

The side length of the given square ABCD = 1

The location of point E = Side $\overline{BC}$ on square ABCD

The location of point F = Side $\overline{CD}$ on square ABCD

∠EAF = 45°

The area of ΔCEF, [CEF] = 1/9 (corrected by using a similar online question)

Required:

To find the value of [AEF]

Solution:

The area of a triangle = (1/2) × Base length × Height

Let x = EC, represent the base length of ΔCEF, and let y = CF represent the height of triangle ΔCEF

We get;

The area of a triangle ΔCEF, [CEF] = (1/2)·x·y = x·y/2

The area of ΔCEF, [CEF] = 1/9 (given)

∴ x·y/2 = 1/9

ΔABE:

$\overline{BE}$ = BC - EC = 1 - x

The area of ΔABE, [ABE] = (1/2)×AB ×BE

AB = 1 = The length of the side of the square

The area of ΔABE, [ABE] = (1/2)× 1 × (1 - x) = (1 - x)/2

ΔADF:

$\overline{DF}$ = CD - CF = 1 - y

The area of ΔADF, [ADF] = (1/2)×AD ×DF

AD = 1 = The length of the side of the square

The area of ΔADF, [ADF] = (1/2)× 1 × (1 - y) = (1 - y)/2

The area of ΔAEF, [AEF] = [ABCD] - [ADF] - [ABE] - [CEF]

[ABCD] = Area of the square = 1 × 1

$[AEF] = 1 - \dfrac{1 - x}{2} - \dfrac{1 - y}{2} - \dfrac{1}{19}= \dfrac{19 \cdot x + 19 \cdot y - 2}{38}$

From $\dfrac{x \cdot y}{2} = \dfrac{1}{9}$ , we have;

$x = \dfrac{2}{9 \cdot y}$ , which gives;

$[AEF] = \dfrac{9 \cdot x + 9 \cdot y - 2}{18}$

Area of a triangle = (1/2) × The product of the length of two sides × sin(included angle between the sides)

∴ [AEF] = (1/2) × $\overline{AE}$ × $\overline{FA}$ × sin(∠EAF)

$\overline{AE}$ = √((1 - x)² + 1), $\overline{FA}$ = √((1 - y)² + 1)

[AEF] = (1/2) × √((1 - x)² + 1) × √((1 - y)² + 1) × sin(45°)

Which by using a graphing calculator, gives;

$\dfrac{1}{2} \times \sqrt{(1 - x)^2 + 1} \times \sqrt{(1 - y)^2 + 1} \times \dfrac{\sqrt{2} }{2} = \dfrac{9 \cdot x + 9 \cdot y - 2}{18}$

Squaring both sides and plugging in $x = \dfrac{2}{9 \cdot y}$ , gives;

$\dfrac{(81 \cdot y^4-180 \cdot y^3 + 200 \cdot y^2 - 40\cdot y +4)\cdot y^2}{324\cdot y^4} = \dfrac{(81\cdot y^4-36\cdot y^3 + 40\cdot y^2 - 8\cdot y +4)\cdot y^2}{324\cdot y^2}$

Subtracting the right hand side from the equation from the left hand side gives;

$\dfrac{40\cdot y- 36\cdot y^2 + 8}{81\cdot y} = 0$

36·y² - 40·y + 8 = 0

$y = \dfrac{40 \pm \sqrt{(-40)^2-4 \times 36\times 8} }{2 \times 36} = \dfrac{5 \pm \sqrt{7} }{9}$

$[AEF] = \dfrac{9 \cdot x + 9 \cdot y - 2}{18} = \dfrac{9 \cdot y^2-2 \cdot y + 2}{18 \cdot y}$

Plugging in $y = \dfrac{5 + \sqrt{7} }{9}$ and rationalizing surds gives;

$[AEF] = \dfrac{9 \cdot \left(\dfrac{5 + \sqrt{7} }{9}\right) ^2-2 \cdot \left(\dfrac{5 + \sqrt{7} }{9}\right) + 2}{18 \cdot \left(\dfrac{5 + \sqrt{7} }{9}\right) } = \dfrac{\dfrac{40+8\cdot \sqrt{7} }{9} }{10+2\cdot \sqrt{7} } = \dfrac{32}{72} = \dfrac{4}{9}$