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3 years ago

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Part A: Factor x2y2 + 6xy2 + 8y2. Show your work. (4 points) Part B: Factor x2 + 8x + 16. Show your work. (3 points) Part C: Fac

tor x2 − 16. Show your work. (3 points) (10 points)

1 answer:

skad [1K]3 years ago

6 0

Answer:

Part A : y²(x + 2)(x + 4)

Part B: (x + 4) (x + 4)

Part C: (x + 4) (x - 4)

Step-by-step explanation:

Part A: Factor x²y²+ 6xy²+ 8y²

x²y²+ 6xy²+ 8y²

y² is very common across the quadratic equation , hence

= y² (x² + 6x + 8)

= (y²) (x² + 6x + 8)

= (y²) (x² + 2x +4x + 8)

= (y²) (x² + 2x)+(4x + 8)

= (y²) (x(x + 2)+ 4(x + 2))

= y²(x+2)(x+4)

Part B: Factor x² + 8x + 16

x² + 8x + 16

= x² + 4x + 4x + 16

= (x² + 4x) + (4x + 16)

= x( x + 4) + 4(x + 4)

= (x + 4) (x + 4)

Part C: Factor x² − 16

= x² − 16

= x² + 0x − 16

= x² + 4x - 4x - 16

= (x² + 4x) - (4x - 16)

= x (x + 4) - 4(x + 4)

= (x + 4) (x - 4)

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3 0

3 years ago

Does anybody know the answer to these questions?

san4es73 [151]

Answer:

1. 625,000 J

2. 100 J

4. 5 kg

5. √5 ≈ 2.236 m/s

Step-by-step explanation:

You should be aware that the SI derived units of Joules are equivalent to kg·m²/s².

To reduce confusion between <em>m</em> for mass and m for meters, we'll use an <em>italic m</em> for mass.

In each case, the "find" variable is what's left after we put the numbers into the formula. It is what the question is asking for. The "given" values are the ones in the problem statement and are the values we put into the formula. The formula is the same in every case.

__

1. KE = (1/2)<em>m</em>v² = (1/2)(2000 kg)(25 m/s)² = 625,000 kg·m²/s² = 625,000 J

__

2. KE = (1/2)<em>m</em>v² = (1/2)(0.5 kg)(20 m/s)² = 100 kg·m²/s² = 100 J

__

4. KE = (1/2)<em>m</em>v²

250 J = (1/2)<em>m</em>(10 m/s)² = 50 m²/s²

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__

5. KE = (1/2)<em>m</em>v²

2000 kg·m²/s² = (1/2)(800 kg)v²

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v = √5 m/s ≈ 2.236 m/s

7 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,

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

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5 0

3 years ago

Read 2 more answers

Help with matrices please? Any wrong/not applicable answers will be reported and BLOCKED

marin [14]

m x H = $\left[\begin{array}{ccc}-25&37.5&-12.5\\\9\end{array}\right]$

Step-by-step explanation:

Step 1; Multiply 5 with this matrix $\left[\begin{array}{ccc}-1&2\\4&8\\\end{array}\right]$ and we get a matrix $\left[\begin{array}{ccc}-5&10\\20&40\\\end{array}\right]$

Multiply the fraction $\frac{2}{5}$ with the matrix $\left[\begin{array}{ccc}-1&2\\4&8\\\end{array}\right]$ and we get $\left[\begin{array}{ccc}-\frac{2m}{5} &\frac{4m}{5} \\\frac{8m}{5} &\frac{16m}{5} \\\end{array}\right]$

Step2; Now equate corresponding values of the matrices with each other.

-5 = $\frac{-2m}{5}$ and so on. By equating we get the value of m as $\frac{25}{2}$

Step 3; Add the matrices to get the value of matrix m.

Adding the three matrices on the RHS we get $\left[\begin{array}{ccc}2&9&-9\\\end{array}\right]$ .

Step 4; Adding the matrices on the LHS we get the resulting matrix as H +

$\left[\begin{array}{ccc}4&6&-8\\\9\end{array}\right]$ . Equating the matrices from step 3 and 4 we get the value of H as $\left[\begin{array}{ccc}-2&3&-1\\\9\end{array}\right]$

Step 5; Now to find the value of m x H we need to multiply the value of $\frac{25}{2}$ with the matrix $\left[\begin{array}{ccc}-2&3&-1\\\9\end{array}\right]$

Step 6; Multiplying we get the matrix m x H = [ -25 $\frac{75}{2}$ $\frac{-25}{2}$ ]

8 0

3 years ago

The time you have before finance charges are assessed is called the a. credit limit. c. APR. b. incentive. d. grace period.

Tanya [424]

Answer:

The time you have before finance charges are assessed is called the Grace period

option-D

Step-by-step explanation:

we know that

Grace period is the extra time given to customer to pay amount before finance charges

But once grace period passes , customer will have to pay extra fee or penalty with interest with passing days

So,

The time you have before finance charges are assessed is called the Grace period

So,

option-D

6 0

4 years ago