All categories

3 years ago

What are the roots of x in - 10x2 + 12x - 9 = 0?

Mathematics

2 answers:

Leya [2.2K]3 years ago

8 0

Answer:

Thus, The two roots of the given quadratic equation are : x = 0.5225 or x = -1.7225

Step-by-step explanation:

The quadratic equation is given to be :

10x² + 12x - 9 = 0

A = 10 , B = 12 , C = 9

Now, using the discriminant rule :

$x=\frac{-B\pm\sqrt{B^2-4AC}}{2A}\\\\\implies x=\frac{-12\pm\sqrt{144+360}}{20}\\\\ \implies x =\frac{-12\pm 22.45}{20}$

So, x = 0.5225 or x = -1.7225

Thus, The two roots of the given quadratic equation are : x = 0.5225 or x = -1.7225

ivann1987 [24]3 years ago

3 0

SORRY FOR CUTTING AND HANDWRITING

HOPE THIS WILL HELP YOU

You might be interested in

Give one number that makes the inequality 3x < 25 true

elena55 [62]

1 is the safest answer.

4 0

3 years ago

Read 2 more answers

At Marco's school, 5/8 of the students are in the band. What percent of the students are in the band?

Harman [31]

62.5%

5/8 = 0.625 then multiply by 100% to get in terms of percent of 62.5%

7 0

2 years ago

A car averaged 48 miles per hour traveling 672 miles from Washington to Orlando. How long was the trip?

lora16 [44]

14 hours
672 miles / 48 mph = 14 hours total

4 0

3 years ago

Read 2 more answers

Find the Maclaurin series for f(x) using the definition of a Maclaurin series. [Assume that f has a power series expansion. Do n

aliya0001 [1]

Answer:

$f(x)=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}$

Step-by-step explanation:

The Maclaurin series of a function f(x) is the Taylor series of the function of the series around zero which is given by

$f(x)=f(0)+f^{\prime}(0)x+f^{\prime \prime}(0)\dfrac{x^2}{2!}+ ...+f^{(n)}(0)\dfrac{x^n}{n!}+...$

We first compute the n-th derivative of $f(x)=\ln(1+2x)$ , note that

$f^{\prime}(x)= 2 \cdot (1+2x)^{-1}\\f^{\prime \prime}(x)= 2^2\cdot (-1) \cdot (1+2x)^{-2}\\f^{\prime \prime}(x)= 2^3\cdot (-1)^2\cdot 2 \cdot (1+2x)^{-3}\\...\\\\f^{n}(x)= 2^n\cdot (-1)^{(n-1)}\cdot (n-1)! \cdot (1+2x)^{-n}\\$

Now, if we compute the n-th derivative at 0 we get

$f(0)=\ln(1+2\cdot 0)=\ln(1)=0\\\\f^{\prime}(0)=2 \cdot 1 =2\\\\f^{(2)}(0)=2^{2}\cdot(-1)\\\\f^{(3)}(0)=2^{3}\cdot (-1)^2\cdot 2\\\\...\\\\f^{(n)}(0)=2^n\cdot(-1)^{(n-1)}\cdot (n-1)!$

and so the Maclaurin series for f(x)=ln(1+2x) is given by

$f(x)=0+2x-2^2\dfrac{x^2}{2!}+2^3\cdot 2! \dfrac{x^3}{3!}+...+(-1)^{(n-1)}(n-1)!\cdot 2^n\dfrac{x^n}{n!}+...\\\\= 0 + 2x -2^2 \dfrac{x^2}{2!}+2^3\dfrac{x^3}{3!}+...+(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}+...\\\\=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^n\dfrac{x^n}{n}$

3 0

3 years ago

Find the surface area of the prism.

alukav5142 [94]

Answer:

A: 6.5 ft

Step-by-step explanation:

Hopefully this helps

4 0

3 years ago

Read 2 more answers

What are the roots of x in - 10x2 + 12x - 9 = 0?​

What are the roots of x in - 10x2 + 12x - 9 = 0?