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4 years ago

Factor 5k 2 - 35k 3. A. 5k2(1 - 7k) B. 5k3(1 - 7k) C. 5k(1 - 7k2)

Mathematics

2 answers:

solniwko [45]4 years ago

4 0

Answer:

The correct answer is A) 5k^2(1 - 7k)

Step-by-step explanation:

To factor, take out the greatest common multiple. Each term has a factor of 5 that can be taken out as a constant and k^2 as a variable. Take both of these out and divide each term by that expression.

5k^2(1 - 7k)

Elanso [62]4 years ago

4 0

its A) 5k^2(1 - 7k) :)

You might be interested in

Q2) for both of the following, calculate A5 and A100. (i) An= 6n-3

dlinn [17]

Answer:

27 and 597

Step-by-step explanation:

To evaluate A5 and A100, substitute n = 5, n = 100 into An , that is

A5 = 6(5) - 3 = 30 - 3 = 27

A100 = 6(100) - 3 = 600 - 3 = 597

4 0

3 years ago

Read 2 more answers

De acuerdo con la tercera ley de movimiento planetario de Kepler, la masa de un planeta es directamente proporcional al cubo de

Sunny_sXe [5.5K]

Answer:

La masa del Sol es $2.509\times 10^{31}$ kilogramos.

Step-by-step explanation:

Tras una lectura cuidadosa al enunciado, tenemos que la Tercera Ley de Kepler queda descrita por la siguiente relación:

$M \propto \frac{r^{3}}{T^{2}}$

$M = k\cdot \frac{r^{3}}{T^{2}}$ (Eq. 1)

Donde:

$r$ - Distancia entre los centros del planeta y el satélite, medido en kilómetros.

$T$ - Período oribital del satélite, medido en días.

$k$ - Constante de proporcionalidad, medida en kilogramo-días cuadrados por kilómetro cúbico.

$M$ - Masa del planeta, medida en kilogramos.

Podemos obtener la masa del Sol mediante la siguiente relación:

$\frac{M_{S}}{M_{E}} = \frac{\frac{r_{E}^{3}}{T_{E}^{2}} }{\frac{r_{M}^{3}}{T_{M}^{2}} }$

$\frac{M_{S}}{M_{E}} = \left(\frac{T_{M}}{T_{E}} \right)^{2}\cdot \left(\frac{r_{E}}{r_{M}} \right)^{3}$ (Eq. 2)

Donde:

$T_{M}$ , $T_{E}$ - Períodos orbitales de la Luna y la Tierra, medidos en días.

$r_{E}$ , $r_{M}$ - Distancias entre la Tierra y el Sol, así como entre la Luna y la Tierra, medidas en kilómetros.

$M_{S}$ , $M_{E}$ - Masas del Sol y la Tierra, medidos en kilogramos.

Si $M_{E} = 75.97\times 10^{24}\,kg$ , $T_{E} = 365.3\,d$ , $T_{M} = 27.3\,d$ , $r_{M} = 3.84\times 10^{5}\,km$ y $r_{E} = 1.496\times 10^{8}\,km$ , entonces tenemos que la masa del Sol es:

$M_{S} = \left(\frac{T_{M}}{T_{E}} \right)^{2}\cdot \left(\frac{r_{E}}{r_{M}} \right)^{3}\cdot M_{E}$

$M_{S} = \left(\frac{27.3\,d}{365.3\,d} \right)^{2}\cdot \left(\frac{1.496\times 10^{8}\,km}{3.84\times 10^{5}\,km} \right)^{3}\cdot (75.97\times 10^{24}\,kg)$

$M_{S} = 2.509\times 10^{31}\,kg$

La masa del Sol es $2.509\times 10^{31}$ kilogramos.

7 0

3 years ago

There are 500 students in a high school senior class. Of these 500 students, 300 regularly wear a necklace to

andre [41]

Answer:

1. P(N) = 3/5

2. P(R) = 2/5

3. P(N and R) = 1/4

4. P(N or R) = 3/4

Step-by-step explanation:

1. Let's review the information given to us to answer the problem correctly:

Number of students in a high school senior class = 500

Number of students that regularly wear a necklace = 300

Number of students that regularly wear a ring = 200

Number of students that wear a necklace and a ring = 125

2. What is P(N), the probability that a senior wears a necklace?

Let's recall the formula of probability:

Probability = Number of favorable outcomes/Total of possible outcomes

Substituting with the real given values:

P(N) = 300/500 = 3/5 (Dividing by 100 numerator and denominator)

P(N) = 3/5

3. What is P(R), the probability that a senior wears a ring?

Substituting with the real given values:

P(R) = 200/500 = 2/5 (Dividing by 100 numerator and denominator)

P(R) = 2/5

4. What is P(N and R), the probability that a senior wears a necklace and a ring?

Substituting with the real given values:

P(N and R) = 125/500 = 1/4 (Dividing by 125 numerator and denominator)

P(N and R) = 1/4

5. What is P(N or R), the probability that a senior wears a necklace or a ring?

P(N or R) = P(N) + P(R) - P(N and R)

Substituting with the real given values:

P(N or R) = 3/5 + 2/5 - 1/4

P(N or R) = 5/5 - 1/4

P(N or R) = 1 - 1/4

P(N or R) = 3/4

5 0

3 years ago

Jacob received a deck of cards for his birthday. The length, width, and height of the card box are shown below. What is the volu

ollegr [7]

The question is incomplete :

The height, width and Lenght isn't Given. However, we can create an hypothetical scenario, with a height 6, length 8 and width 4

Answer:

192 unit³

Step-by-step explanation:

The volume of the card box :

Recall the volume of box formula :

V = length * width * height

Volume = 8 * 6 * 4

Volume = 192 unit³

This is the procedure for any given dimension of the card deck.

6 0

3 years ago

Peter attends 6 lessons each week . A year he attends 52 weeks. But he missed 5 classes how many lessons did he attend during th

VMariaS [17]

You are given the information that he takes six lessons per week. First we will calculate the total lessons he could potentially take per year.

6 • 52 = 312

He could potentially take 312 lessons in one year.

Now that you know this information, you simply subtract the days he missed from that total.

312 - 5 = 307

Your final answer: Peter took 307 lessons during the year.

4 0

3 years ago