All categories

3 years ago

(-2.6) - 3.8 Help me plsss

Mathematics

1 answer:

natali 33 [55]3 years ago

7 0

Answer:

-6.4

Step-by-step explanation:

it's the answer trust me

You might be interested in

Plz tell me the answer

solong [7]

medium box has best value

8 0

3 years ago

1) Determine the discriminant of the 2nd degree equation below:

Aleksandr-060686 [28]

$\LARGE{ \boxed{ \mathbb{ \color{purple}{SOLUTION:}}}}$

We have, Discriminant formula for finding roots:

$\large{ \boxed{ \rm{x = \frac{ - b \pm \: \sqrt{ {b}^{2} - 4ac} }{2a} }}}$

Here,

x is the root of the equation.
a is the coefficient of x^2
b is the coefficient of x
c is the constant term

1) Given,

3x^2 - 2x - 1

Finding the discriminant,

➝ D = b^2 - 4ac

➝ D = (-2)^2 - 4 × 3 × (-1)

➝ D = 4 - (-12)

➝ D = 4 + 12

➝ D = 16

2) Solving by using Bhaskar formula,

❒ p(x) = x^2 + 5x + 6 = 0

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 5\pm \sqrt{( - 5) {}^{2} - 4 \times 1 \times 6 }} {2 \times 1}}}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 5 \pm \sqrt{25 - 24} }{2 \times 1} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 5 \pm 1}{2} }}$

So here,

$\large{\boxed{ \rm{ \longrightarrow \: x = - 2 \: or - 3}}}$

❒ p(x) = x^2 + 2x + 1 = 0

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 2 \pm \sqrt{ {2}^{2} - 4 \times 1 \times 1} }{2 \times 1} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 2 \pm \sqrt{4 - 4} }{2} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 2 \pm 0}{2} }}$

So here,

$\large{\boxed{ \rm{ \longrightarrow \: x = - 1 \: or \: - 1}}}$

❒ p(x) = x^2 - x - 20 = 0

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - ( - 1) \pm \sqrt{( - 1) {}^{2} - 4 \times 1 \times ( - 20) } }{2 \times 1} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{ 1 \pm \sqrt{1 + 80} }{2} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{1 \pm 9}{2} }}$

So here,

$\large{\boxed{ \rm{ \longrightarrow \: x = 5 \: or \: - 4}}}$

❒ p(x) = x^2 - 3x - 4 = 0

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - ( - 3) \pm \sqrt{( - 3) {}^{2} - 4 \times 1 \times ( - 4) } }{2 \times 1} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{3 \pm \sqrt{9 + 16} }{2 \times 1} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{3 \pm 5}{2} }}$

So here,

$\large{\boxed{ \rm{ \longrightarrow \: x = 4 \: or \: - 1}}}$

<u>━━━━━━━━━━━━━━━━━━━━</u>

5 0

3 years ago

Read 2 more answers

Use the Rational Zeros Theorem to write a list of all potential rational zeros f(x) = 14x^3 + 56x^2 + 2x - 7

Crank

The factors of 7are -1 and 7 or 1 and -7, the factors of 14 are 1, 2, 7, and 14, or -1, -2, -7,-14. so the list of potential zeros are: 1/1, 1/2, 1/7, 1/14, 7/1,7/2, 7/7, 7/14, which can be simplified into 1, 1/2,1/7, 1/14, 7, 7/2
add the negative ones: -1, -1/2,-1/7, -1/14, -7, -7/2
I believe there are a total of 12 potential zeros
reference:
http://www.sparknotes.com/math/algebra2/polynomials/section4.rhtml

3 0

3 years ago

Select the equation written in point-slope form.

Kisachek [45]

D, y-y1=m(x-x1)
^^^^^^^^^^^^^^^

6 0

3 years ago

7*10*2+5<br> HELP HELP HELPPPPPPP

Inessa [10]

$7 \times 10 \times 2 + 5 \\ \\ 140 + 5 \\ \\ 145 \\ \\ Answer: \fbox {145}$

6 0

3 years ago

Read 2 more answers