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3 years ago

Tricia listed her assets and liabilities.

Mathematics

2 answers:

kumpel [21]3 years ago

6 0

Answer:

<u>The correct answer is A. Credit card bill and B. Car loan</u>

Step-by-step explanation:

Let's recall that we consider a personal or familiar liability an obligation to pay money you or your family owe to another person or financial institution like a bank. Liabilities can be classified according to the term, either short such as a credit card or loan from the bank with one-year term, or long such as a car loan or mortgage.

According to that definition, Tricia's liabilities are A. Credit card bill and B. Car loan.

lesantik [10]3 years ago

6 0

Answer:

its a and b

Step-by-step explanation:

You might be interested in

What is the binomial expansion of (x 2y)7? 2x7 14x6y 42x5y2 70x4y3 70x3y4 42x2y5 14xy6 2y7 x7 14x6y 42x5y2 70x4y3 70x3y4 42x2y5

AysviL [449]

You can take x = a, 2y = b and then can apply the binomial theorem.

The expansion of given expression is given by:

Option D: $x^7 + 14x^6y + 84x^5y^2 + 280x^4y^3 + 560x^3y^4 + 672x^2y^5 + 448xy^6 + 128y^7$ is

<h3>What is binomial theorem?</h3>

It provides algebraic expansion of exponentiated(integer) binomial.

According to binomial theorem,

$(a+b)^n = \sum_{i=0}^n ^nC_i a^ib^{n-i}$

<h3>How to use binomial theorem for given expression?</h3>

Taking a = x, and b =2y, we have n = 7, thus:

$(x+2y)^7 = \: ^7C_0x^0(2y)^7 + \: ^7C_1x^1(2y)^6 + \: ^7C_2x^2(2y)^5 + \: ^7C_3x^3(2y)^4 + \:^7C_4x^4(2y)^3 + \:^7C_5x^5(2y)^2 + \:^7C_6x^6y^1 + \: ^7C_0x^7y^0\\\\
(x+2y)^7 = 128y^7 + 448xy^6 + 672x^2y^5 + 560x^3y^4 + 280x^4y^3 + 84x^5y^2 + 14x^6y + x^7\\\\
(x+2y)^7 = x^7 + 14x^6y + 84x^5y^2 + 280x^4y^3 + 560x^3y^4 + 672x^2y^5 + 448xy^6 + 128y^7$

Thus, Option D: $x^7 + 14x^6y + 84x^5y^2 + 280x^4y^3 + 560x^3y^4 + 672x^2y^5 + 448xy^6 + 128y^7$ is correct.

Learn more about binomial theorem here:

brainly.com/question/86555

6 0

3 years ago

Just make sure to help me out when I need it. Thx!

Mnenie [13.5K]

Answer:

okay i will

Step-by-step explanation:

5 0

3 years ago

In a circus performance, a monkey is strapped to a sled and both are given an initial speed of 3.0 m/s up a 22.0° inclined track

Aloiza [94]

Answer:

Approximately $0.31\; \rm m$ , assuming that $g = 9.81\; \rm N \cdot kg^{-1}$ .

Step-by-step explanation:

Initial kinetic energy of the sled and its passenger:

$\begin{aligned}\text{KE} &= \frac{1}{2}\, m \cdot v^{2} \\ &= \frac{1}{2} \times 14\; \rm kg \times (3.0\; \rm m\cdot s^{-1})^{2} \\ &= 63\; \rm J\end{aligned}$ .

Weight of the slide:

$\begin{aligned}W &= m \cdot g \\ &= 14\; \rm kg \times 9.81\; \rm N \cdot kg^{-1} \\ &\approx 137\; \rm N\end{aligned}$ .

Normal force between the sled and the slope:

$\begin{aligned}F_{\rm N} &= W\cdot \cos(22^{\circ}) \\ &\approx 137\; \rm N \times \cos(22^{\circ}) \\ &\approx 127\; \rm N\end{aligned}$ .

Calculate the kinetic friction between the sled and the slope:

$\begin{aligned} f &= \mu_{k} \cdot F_{\rm N} \\ &\approx 0.20\times 127\; \rm N \\ &\approx 25.5\; \rm N\end{aligned}$ .

Assume that the sled and its passenger has reached a height of $h$ meters relative to the base of the slope.

Gain in gravitational potential energy:

$\begin{aligned}\text{GPE} &= m \cdot g \cdot (h\; {\rm m}) \\ &\approx 14\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \times h\; {\rm m} \\ & \approx (137\, h)\; {\rm J} \end{aligned}$ .

Distance travelled along the slope:

$\begin{aligned}x &= \frac{h}{\sin(22^{\circ})} \\ &\approx \frac{h\; \rm m}{0.375}\end{aligned}$ .

The energy lost to friction (same as the opposite of the amount of work that friction did on this sled) would be:

$\begin{aligned} & - (-x)\, f \\ = \; & x \cdot f \\ \approx \; & \frac{h\; {\rm m}}{0.375}\times 25.5\; {\rm N} \\ \approx\; & (68.1\, h)\; {\rm J}\end{aligned}$ .

In other words, the sled and its passenger would have lost (approximately) $((137 + 68.1)\, h)\; {\rm J}$ of energy when it is at a height of $h\; {\rm m}$ .

The initial amount of energy that the sled and its passenger possessed was $\text{KE} = 63\; {\rm J}$ . All that much energy would have been converted when the sled is at its maximum height. Therefore, when $h\; {\rm m}$ is the maximum height of the sled, the following equation would hold.

$((137 + 68.1)\, h)\; {\rm J} = 63\; {\rm J}$ .

Solve for $h$ :

$(137 + 68.1)\, h = 63$ .

$\begin{aligned} h &= \frac{63}{137 + 68.1} \approx 0.31\; \rm m\end{aligned}$ .

Therefore, the maximum height that this sled would reach would be approximately $0.31\; \rm m$ .

7 0

2 years ago

:(((( i can't get this one! Please help...

ValentinkaMS [17]

Answer:

4877 fish

Step-by-step explanation:

Each year, the fish population is multiplied by 1-6% = 0.94, so after 8 years it has been multiplied by 0.94^8 ≈ 0.609569.

At that time, the population is ...

8000×0.609569 ≈ 4877 . . . fish

6 0

3 years ago

Find the radius and diameter of circle A, if the endpoints of the radius<br> are (-2, -1) and (6,4).

AleksandrR [38]

Step-by-step explanation:

$length \: between \: ( - 2, - 1) \: and \: (6,4) \\ = \sqrt{(x _{1} - x _{2} )² + (y _{1} - y _{2})² } \\ = \sqrt{( - 2 - 6)²+ ( - 1 - 4)²} \\ = \sqrt{64 + 25} \\ = \sqrt{100} \\ = 10 \: units \\ radius = 10 \: units \\ diameter = 2 \times radius \\ d = 2 \times 10 \\ d = 20 \: units$

8 0

3 years ago