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3 years ago

Which expression is equal to a•a•a•a•b•b

Mathematics

2 answers:

Alona [7]3 years ago

8 0

A*a*a*a*b*b simplifies into a^4*b^2 or $a^{4}$ $b^{2}$

Ronch [10]3 years ago

8 0

Since there are 4 A's and 2 B's the expression would be
a^4b^2

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What is the value of n?

melamori03 [73]

The answer for this question that needs an answer is n = 5

because 18(n-1) = 9(n+7) implies that 18n - 18 = 9n + 63 which implies that n =5 for the equation to hold true.

6 0

3 years ago

The area of a rectangular barnyard is given by the trinomial 6x2 + 7x – 20. What are the possible dimensions of the barnyard? Us

sertanlavr [38]

Area of a rectangular barnyard: A=6x^2+7x-20
A=6(6x^2+7x-20)/6
A=[(6^2)(x^2)+7(6x)-120]/6
A=[(6x)^2+7(6x)-120]/6
A=(6x+15)(6x-8)/6
A=(6x+15)(6x-8)/[(3)(2)]
A=(6x/3+15/3)(6x/2-8/2)
A=(2x+5)(3x-4)
A=bh
b=2x+5; h=3x-4
Answer: The possible dimensions of the barnyard are 2x+5 and 3x-4

8 0

2 years ago

2 women and 7 girls can do a peice of work in 4 days.4 women and 4 girls can do the same work in 3 days. How long would it take

svetlana [45]

Answer:

x: work a woman can do in 1 day

y: work a girl can do in 1 day

Then

2x + 7y =1/4

4x + 4y =1/3

4x + 14y =2/4 (1)

4x + 4y =1/3 (2)

=> Let (1) - (2), 10y =2/4-1/3 <=>10y = 1/6 <=> y = 1/(6x10) = 1/60 (work)

=> From (2),4x = 1/3 - 4x1/60 => 4x = 16/60 => x = 4/60 (work)

=> 1 day, a woman and a girl can do: 1/60+4/60 = 5/60 =1/12 (work)

Then, the day required for a woman and a girl to complete work: 12 days

3 0

3 years ago

A car starts with a dull tank of gas. After driving around 5he city, 1/7 of the gas has been used. With the rest of the gas in t

Sati [7]

Answer:

$\frac{2}{7}$

Step-by-step explanation:

Given:

A car starts with a dull tank of gas

1/7 of the gas has been used around the city.

With the rest of the gas in the car, the car can travel to and from Ottawa three times.

Question asked:

What fractions of a tank of gas does each complete trip to Ottawa use?

Solution:

Fuel used around the city = $\frac{1}{7}$

Remaining fuel after driving around the city = 1 - $\frac{1}{7}$

= $\frac{7 - 1}{7} = \frac{6}{7}$

According to question:

As from the rest of the gas in the car that is $\frac{6}{7}$ , the car can complete 3 trip to Ottawa which means,

By unitary method:

The car can complete 3 trip by using = $\frac{6}{7}$ tank of gas.

The car can complete 1 trip by using = $\frac{6}{7} \div 3$

= $\frac{6}{7} \times\frac{1}{3}$

= $\frac{6}{21}$

= $\frac{2}{7}$ tank of gas

Thus, $\frac{2}{7}$ tank of gas used for each complete trip to Ottawa.

5 0

3 years ago

Find the numbers b such that the average value of f(x) = 7 + 10x − 9x2 on the interval [0, b] is equal to 8.

barxatty [35]

Answer:

The numbers $b$ such that the average value of $f(x) = 7 +10\cdot x - 9\cdot x^{2}$ on the interval [0, b] is equal to 8 are $b_{1} \approx 1.434$ and $b_{2} \approx 0.232$ .

Step-by-step explanation:

The mean value of function within a given interval is given by the following integral:

$\bar f = \frac{1}{b-a}\cdot \int\limits^b_a {f(x)} \, dx$

If $f(x) = 7 +10\cdot x - 9\cdot x^{2}$ , $a = 0$ , $b = b$ and $\bar f = 8$ , then:

$\frac{1}{b}\cdot \int\limits^b_0 {7+10\cdot x -9\cdot x^{2}} \, dx = 8$

$\frac{7}{b}\int\limits^b_0 \, dx + \frac{10}{b} \int\limits^b_0 {x}\, dx - \frac{9}{b} \int\limits^b_0 {x^{2}}\, dx = 8$

$\left(\frac{7}{b} \right)\cdot b + \left(\frac{10}{b} \right)\cdot \left(\frac{b^{2}}{2} \right)-\left(\frac{9}{b} \right)\cdot \left(\frac{b^{3}}{3} \right) = 8$

$7 + 5\cdot b - 3\cdot b^{2} = 8$

$3\cdot b^{2}-5\cdot b +1 = 0$

The roots of this polynomial are determined by the Quadratic Formula:

$b_{1} \approx 1.434$ and $b_{2} \approx 0.232$ .

The numbers $b$ such that the average value of $f(x) = 7 +10\cdot x - 9\cdot x^{2}$ on the interval [0, b] is equal to 8 are $b_{1} \approx 1.434$ and $b_{2} \approx 0.232$ .

7 0

3 years ago