All categories

4 years ago

Consider the quadratic equation below. Determine the correct set-up for solving the equation using the quadratic formula.

Mathematics

1 answer:

Studentka2010 [4]4 years ago

5 0

Answer:

--3 ±sqrt((-3)^2 -4(4)(-9))

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

2(4)

Step-by-step explanation:

4x^2-5=3x+4

We need to get this in standard form to answer the question

Subtract 3x from each side

4x^2-3x-5=3x-3x+4

4x^2-3x-5=+4

Subtract 4 from each side

4x^2-3x-9 =0

a = 4

b = -3

c = -9

-b ±sqrt(b^2 -4ac)

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

--3 ±sqrt((-3)^2 -4(4)(-9))

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

2(4)

You might be interested in

Solve 3 x cube minus 4 x square

motikmotik

Answer:

The answer is 3x^3-4x^2

7 0

3 years ago

Please i need an answer urgently. Sum of 3 terms are 252, first term is 192. Find common ratio

Viefleur [7K]

Answer:

$r=\frac{1}{4}\\\\r=-\frac{5}{4}$

Step-by-step explanation:

<u>Geometric Sequences</u>

The geometric sequence is given as:

$a_1, a_1r, a_1r^2, a_1r^3,..., a_1r^{n-1}$

Where n is the number of the term, n≥1, and r is the common ratio.

The sum of n terms of the geometric sequence is given by:

$\displaystyle S_n=a_1\frac{r^n-1}{r-1}$

We are given: S3=252, a1=192, thus substuting:

$\displaystyle 252=192\frac{r^3-1}{r-1}$

Dividing by 12:

$\displaystyle 21=16\frac{r^3-1}{r-1}$

Recall that:

$r^3-1=(r-1)(r^2+r+1)$

Substituting:

$\displaystyle 21=16\frac{(r-1)(r^2+r+1)}{r-1}$

Simplifying:

$21=16(r^2+r+1)=16r^2+16r+16$

Rearranging:

$16r^2+16r+16-21=0$

16r^2+16r-5=0

Rewriting:

$16r^2-4r+20r-5=0$

Factoring:

$4r(4r-1)+5(4r-1)=0$

$(4r-1)(4r+5)=0$

Solving:

$r=\frac{1}{4}\\\\r=-\frac{5}{4}$

Both solutions are valid

6 0

3 years ago

I keep getting exactly 50%! Help!

Y_Kistochka [10]

Total = 242 + 307 = 549 but 242 ordered salad
so
242 / 549 = 0.4408 = 44.08%

answer
C. 44.08%

6 0

3 years ago

Read 2 more answers

Help me with geometry. I don’t understand this

valkas [14]

Answer:

b and d are 176 c is 53 a is 51

5 0

3 years ago

Read 2 more answers

What is the equation of the following line? Be sure to scroll down first to see

likoan [24]

$\bf (\stackrel{x_1}{-3}~,~\stackrel{y_1}{1})\qquad (\stackrel{x_2}{0}~,~\stackrel{y_2}{0}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{0}-\stackrel{y1}{1}}}{\underset{run} {\underset{x_2}{0}-\underset{x_1}{(-3)}}}\implies \cfrac{-1}{0+3}\implies -\cfrac{1}{3}$

$\bf \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{1}=\stackrel{m}{-\cfrac{1}{3}}[x-\stackrel{x_1}{(-3)}]\implies y-1=-\cfrac{1}{3}(x+3) \\\\\\ y-1=-\cfrac{1}{3}x-1\implies y=-\cfrac{1}{3}x$

3 0

4 years ago