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3 years ago

Please find the value of x

Mathematics

1 answer:

Verizon [17]3 years ago

6 0

I am not sure if this is right , but I tried . I hope I helped !

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A geometric transformation that does not change the shape or size of the geometric object is called a

Pavlova-9 [17]

Answer:

Univ.

Explanation:

A geometry transformation is either rigid or non-rigid; another word for a rigid transformation is "isometry". An isometry, such as a rotation, translation, or reflection, does not change the size or shape of the figure.

8 0

3 years ago

What is the value of y in the equation 9 + y = 8? this is extremely time sensitive

adoni [48]

You have to switch the 8 to the left and the y to the right which would change the equation to 9-8=y

y=1

5 0

3 years ago

Read 2 more answers

What’s is the solution to the equation x to the power of 2 =16/25

skelet666 [1.2K]

Answer:

$x=\dfrac{4}{5}$

Step-by-step explanation:

The given equation is :

$x^2=\dfrac{16}{25}$

We need to find the value of x.

When the exponent 2 removes from LHS, it will shift to RHS as square root as follows :

$x=\sqrt{\dfrac{16}{25}}$

We know that, 4×4= 16 and 5×5 = 25

So,

$x=\sqrt{\dfrac{4\times 4}{5\times 5}}\\\\x=\dfrac{4}{5}$

So, the solution of the given equation is 4/5.

4 0

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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5 0

3 years ago

Read 2 more answers

Help me please.....:)

kozerog [31]

Answer:I don’t know if I’m right but imma have to say G

Step-by-step explanation:

I’m not good at math

8 0

3 years ago