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3 years ago

Please help me and my daughter.

Mathematics

2 answers:

valina [46]3 years ago

7 0

Step-by-step explanation:

3x2−7x−20=03x2-7x-20=0

Use the quadratic formula to find the solutions.

−b±√b2−4(ac)2a-b±b2-4(ac)2a

Substitute the values a=3a=3, b=−7b=-7, and c=−20c=-20 into the quadratic formula and solve for xx.

7±√(−7)2−4⋅(3⋅−20)2⋅37±(-7)2-4⋅(3⋅-20)2⋅3

Simplify.

Tap for more steps...

x=7±176x=7±176

The final answer is the combination of both solutions.

x=4,−53

garri49 [273]3 years ago

3 0

Answer:

you can either factorise or use tge formula method

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14. In a right triangle, if one acute angle has measure 10x + 6 and the other has measure 2x + 7, find the larger of these two a

lubasha [3.4K]

Answer:

421/6 degrees

Step-by-step explanation:

The acute angles of a right triangle are complementary, so

$10x+6+2x+7=90 \\ \\ 12x+13=90 \\ \\ 12x=77 \\ \\ x=\frac{77}{12}$

So, the acute angles measures 421/6 degrees and 119/6 degrees, and thus the larger angle is 421/6 degrees.

7 0

1 year ago

Solve for the variable: 2x - 28 = 16

svlad2 [7]

2x = 44
44/2 = x
X= 22

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2 years ago

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Likurg_2 [28]

Answer:

1/2

Step-by-step explanation:

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2 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,

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

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2 years ago

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IrinaVladis [17]

Answer:

C. 2144

Step-by-step explanation:

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2 years ago

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