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4 years ago

Can you please help me. When you answer can you show work on piece of paper and take picture.

Mathematics

1 answer:

antoniya [11.8K]4 years ago

3 0

$\bf \textit{difference of squares}
\\ \quad \\
(a-b)(a+b) = a^2-b^2\qquad \qquad 
a^2-b^2 = (a-b)(a+b)\\\\
-------------------------------\\\\$

$\bf \cfrac{x^2}{x^2-4}-\cfrac{x+1}{x+2}\implies \cfrac{x^2}{x^2-2^2}-\cfrac{x+1}{x+2}\implies \cfrac{x^2}{(x-2)(x+2)}-\cfrac{x+1}{x+2}
\\\\\\
\textit{so our LCD is then }(x-2)(x+2)
\\\\\\
\cfrac{[x^2]~~-~~[(x+1)(x-2)]}{(x-2)(x+2)}\implies \cfrac{[x^2]~~-~~[x^2-x-2]}{(x-2)(x+2)}
\\\\\\
\cfrac{[\underline{x^2}]~~\underline{-x^2}+x+2}{(x-2)(x+2)}\implies \cfrac{\underline{x+2}}{(x-2)\underline{(x+2)}}\implies \cfrac{1}{x-2}$

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Find any domain restrictions on the given rational equation:

koban [17]

Answer: 10, -10

Step-by-step explanation: a pex

3 0

3 years ago

The function f(x)f(x) is a quartic function and the zeros of f(x)f(x) are -6−6, -5−5, -2−2 and 11. Assume the leading coefficien

Luden [163]

<h2>Answer: </h2>

$\boxed{f(x) = 11x^4 + 22x^3 - 1001x^2 - 5632x - 7260}$

<h2>Explanation: </h2>

A quartic function is a function given by the the following equation in standard form:

$f(x)=a_{4}x^4+a_{3}x^3+a_{2}x^2+a_{1}x+a_{0} \\ \\ \\ Where: \\ \\ a_{4},a_{3},a_{2},a_{1},a_{0} \ \text{are constant} \\ \\ \text{and} \ a_{4} \ \text{is the leading coefficient}$

From the statement we must assume the leading coefficient of f(x) is 11, so:

$a_{4}=11$

The zeros are:

$x=-6 \\ \\ x=-5 \\ \\ x=-2 \\ \\ x=11$

So, we can write:

$f(x)=11(x-(-6))(x-(-5))(x-(-2))(x-11) \\ \\ \\ So: \\ \\ \mathrm{Rule}:-\left(-a\right)=a \\ \\ \\ Then: \\ \\ f(x)=11\left(x+6\right)\left(x+5\right)\left(x+2\right)\left(x-11\right)$

$Expand:\left(x+6\right)\left(x+5\right):\ x^2+11x+30: \ \text{Distributive Property} \\ \\ \\ Then: \\ \\ f(x)=11\left(x^2+11x+30\right)\left(x+2\right)\left(x-11\right) \\ \\ \\ Expand:\left(x^2+11x+30\right)\left(x+2\right):\ x^3+13x^2+52x+60: \ \text{Distributive Property} \\ \\ \\ Then: \\ \\ f(x)=11\left(x^3+13x^2+52x+60\right)\left(x-11\right)$

$Expand: \left(x^3+13x^2+52x+60\right)\left(x-11\right):\ x^4+2x^3-91x^2-512x-660 \\ (Distributive \ property)$

Finally:

$f(x) = 11(x^4 + 2x^3 - 91x^2 - 512x - 660) \\ \\ \boxed{f(x) = 11x^4 + 22x^3 - 1001x^2 - 5632x - 7260}$

3 0

3 years ago

Josh earns $2700 each summer and puts the money in a savings account with a 4.5% simple interest rate. If he deposits the same a

iragen [17]

Answer: Option 'B' is correct.

Step-by-step explanation:

Since we have given that

Amount Josh earn each summer = $2700

Rate of simple interest = 4.5%

According to question, he deposits the same amount each year.

First summer, he deposited = $2700

In Second summer, he deposited = $2700×2= $5400

In Third summer, he deposited = $2700 × 3=$8100

At the end of his third summer, his money becomes,

$\text{Simple Interest}=\frac{P\times R\times T}{100}\\\\\text{Simple Interest}=\frac{8100\times 4.5\times 1}{100}\\\\\text{Simple Interest}=\$364.5$

As we know the formula for "Amount" that it becomes at the end of his third summer:

$Amount=P+S.I\\\\Amount=\$8100+\$364.5\\\\Amount=\$8464.5$

Hence, Option 'B' is correct.

6 0

4 years ago

Find the perimeter<br><br> A) 66.4<br> B) 57.8<br> C) 62.7<br> D)59.2

velikii [3]

BBB it is b the answer is b

4 0

3 years ago

If an arrow is shot upward on Mars with a speed of 62 m/s, its height in meters t seconds later is given by y = 62t − 1.86t². (R

Soloha48 [4]

Answer:

Approximately $58.28\; \rm m \cdot s^{-1}$ .

Step-by-step explanation:

The velocity of an object is the rate at which its position changes. In other words, the velocity of an object is equal to the first derivative of its position, with respect to time.

Note that the arrow here is launched upwards. (Assume that the effect of wind on Mars is negligible.) There would be motion in the horizontal direction. The horizontal position of this arrow will stays the same. On the other hand, the vertical position of this arrow is the same as its height: $y = 62\, t - 1.86\, t^2$ .

Apply the power rule to find the first derivative of this $y$ with respect to time $t$ .

By the power rule:

the first derivative of $t$ (same as
the first derivative of $t^2$ (same as $t$ to the second power) with respect to

Therefore:

$\begin{aligned}\frac{dy}{d t} &= \frac{d}{d t}\left[62 \, t - 1.86\, t^2\right] \\ &= 62\,\left(\frac{d}{d t}\left[t\right]\right) - 1.86\, \left(\frac{d}{d t}\left[t^2\right]\right) \\ &= 62 \times 1 - 1.86\times\left(2\, t) = 62 - 3.72\, t\end{aligned}$ .

In other words, the (vertical) velocity of this arrow at time $t$ would be $(62 - 3.72\, t)$ meters per second.

Evaluate this expression for $t = 1$ to find the (vertical) velocity of this arrow at that moment: $62 - 3.72 \times 1 =58.28$ .

6 0

3 years ago

Read 2 more answers