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2 years ago

(x-1)+(2x+3)

1 answer:

8 0

1) =3x+2
2) =−x−3
3) =7x+45
4) =14x−1

Should I show work also?

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. Shanice is making a garden. The area of the garden is 24 ft2. Use the following shape and information to find the base (6−4) a

crimeas [40]

Answer:

The base and height of the garden are 4.60 ft and 5.60 ft.

Step-by-step explanation:

The area of the garden is, A = 24 ft².

The base height of the garden are:

base (b) = 6x - 4

height (h) = x + 3

Compute the value of x as follows:

$A=\frac{1}{2}\times b \times h\\\\24=\frac{1}{2}\times (6x-4) \times (x+3)\\\\48=6x^{2}+18x-4x-12\\\\6x^{2}+14x-60=0\\\\3x^{2}+7x-30=0\\$

The last equation is a quadratic equation.

Compute the roots of the quadratic equation as follows:

$x = \frac{ -b \pm \sqrt{b^2 - 4ac}}{ 2a }\\x = \frac{ -14 \pm \sqrt{14^2 - 4(3)(-30)}}{ 2(3) }\\x = \frac{ -14 \pm \sqrt{556}}{ 6 }\\x = \frac{ -14 }{ 6 } \pm \frac{2\sqrt{139}\, }{ 6 }\\x = -\frac{ 7}{ 3 } \pm \frac{ \sqrt{139}\, }{ 3 }\\x = 1.59661\\x\approx 1.60$

The value of base and height are:

$\text{base}=6x-4=(6\times 1.60)-4=5.60\ \text{ft}\\\\\text{height}=x+3=1.60+3=4.60\ \text{ft}$

Thus, the base and height of the garden are 4.60 ft and 5.60 ft.

3 0

3 years ago

HELP Which of the lines below has a slope of 0?

Juliette [100K]

Answer:

C has a zero slope

Step-by-step explanation:

A horizontal line has a slope of zero

A vertical line has an undefined slope

A positive slope goes up from left to right

A negative slope goes down from left to right

4 0

3 years ago

What is 5x-15x + 20y = x + 5 Solve for y

Setler79 [48]

Y=11x/20 +.25
Solve for y by simplifying both sides of the equation, then isolating the variable.

4 0

3 years ago

Read 2 more answers

I need help ASAP, would really appreciate it, marking first as brainliest.

swat32

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3 0

2 years ago

Read 2 more answers

Evaluate the difference quotient for the given function. Simplify your answer.

nasty-shy [4]

I suppose you mean

$f(x) = \dfrac{x+5}{x+1}$

Then

$f(3) = \dfrac{3+5}{3+1} = \dfrac84 = 2$

and the difference quotient is

$\dfrac{f(x)-f(3)}{x-3} = \dfrac{\frac{x+5}{x+1}-2}{x-3} \\\\ \dfrac{f(x)-f(3)}{x-3} = \dfrac{\frac{x+5-2(x+1)}{x+1}}{x-3} \\\\ \dfrac{f(x)-f(3)}{x-3} = \dfrac{-x+3}{(x+1)(x-3)} \\\\ \dfrac{f(x)-f(3)}{x-3} = \boxed{-\dfrac{x-3}{(x+1)(x-3)}}$

If it's the case that x ≠ 3, then (x - 3)/(x - 3) reduces to 1, and you would be left with

$\dfrac{f(x)-f(3)}{x-3}\bigg|_{x\neq3} = -\dfrac1{x+1}$

4 0

2 years ago