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3 years ago

Greatest common factor of the following monomials 40a^3b and 24a^3b^5

Mathematics

1 answer:

timofeeve [1]3 years ago

8 0

the gcf is product of gcf each factor,

gcf(40a³b,24a³b⁵)

= gcf(40,24)•gcf(a³,a³)•gcf(b,b⁵)

= 8a³b

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In a young woman's study group, 7 percent of the students are from the town where the young woman's school is located, 80 percen

Umnica [9.8K]

Answer: the percent of study group students from outside the state is 13%

Step-by-step explanation:

The total number of students in the young woman's study group that are from the town where the young woman's school is located is 7%

The total number of students in the young woman's study group that are from in-state (but not from the school's town) is 80%.

Since the rest are from outside the state, the percent of study group students from outside the state would be

100 - (80 + 7) = 100 - 87 = 13%

5 0

3 years ago

Can someone please help me with this

vovangra [49]

Answer:

y = 5

z = 55

Step-by-step explanation:

90 - 35 = 55

z = 55

7y = 35

$\frac{7y}{7} = \frac{35}{7}$

y= 5

3 0

3 years ago

Simplify the expression. (7x + 3) + (4x - 1) - (2x + 4)

Crazy boy [7]

Answer:

9x -2

Step-by-step explanation:

(7x + 3) + (4x - 1) - (2x + 4)

Distribute the minus sign

(7x + 3) + (4x - 1) - 2x - 4

Combine like terms

7x+4x-2x +3 -1-4

9x -2

3 0

3 years ago

214 people used a public swimming pool. $1.50 for kids and $2.25 for adults. The receipt totaled to $390 How many children and h

zysi [14]

Answer:the number of kids that used the public pool that day is 122

the number of adults that used the public pool that day is 92

Step-by-step explanation:

Let x represent the number of kids that used the public pool that day.

Let y represent the number of adults that used the public pool that day.

214 people used a public swimming pool. This means that

x + y = 214

1.50 for kids and $2.25 for adults. The receipt totaled to $390. This means that

1.5x + 2.25y = 390 - - - - - - - - - -1

Substituting x = 214 - y into equation 1, it becomes

1.5x

1.5(214 - y) + 2.25y = 390

321 - 1.5y + 2.25y = 390

- 1.5y + 2.25y = 390 - 321

0.75y = 69

y = 69/0.75 = 92

Substituting y = 92 into x = 214 - y, it becomes

x = 214 - 92 = 122

5 0

3 years ago

1. Express <img src="https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bx%282x%2B3%29%20%7D" id="TexFormula1" title="\frac{1}{x(2x+3) }" a

katovenus [111]

1. Let a and b be coefficients such that

$\dfrac1{x(2x+3)} = \dfrac ax + \dfrac b{2x+3}$

Combining the fractions on the right gives

$\dfrac1{x(2x+3)} = \dfrac{a(2x+3) + bx}{x(2x+3)}$

$\implies 1 = (2a+b)x + 3a$

$\implies \begin{cases}3a=1 \\ 2a+b=0\end{cases} \implies a=\dfrac13, b = -\dfrac23$

so that

$\dfrac1{x(2x+3)} = \boxed{\dfrac13 \left(\dfrac1x - \dfrac2{2x+3}\right)}$

2. a. The given ODE is separable as

$x(2x+3) \dfrac{dy}dx} = y \implies \dfrac{dy}y = \dfrac{dx}{x(2x+3)}$

Using the result of part (1), integrating both sides gives

$\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3|\right) + C$

Given that y = 1 when x = 1, we find

$\ln|1| = \dfrac13 \left(\ln|1| - \ln|5|\right) + C \implies C = \dfrac13\ln(5)$

so the particular solution to the ODE is

$\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3|\right) + \dfrac13\ln(5)$

We can solve this explicitly for y :

$\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3| + \ln(5)\right)$

$\ln|y| = \dfrac13 \ln\left|\dfrac{5x}{2x+3}\right|$

$\ln|y| = \ln\left|\sqrt[3]{\dfrac{5x}{2x+3}}\right|$

$\boxed{y = \sqrt[3]{\dfrac{5x}{2x+3}}}$

2. b. When x = 9, we get

$y = \sqrt[3]{\dfrac{45}{21}} = \sqrt[3]{\dfrac{15}7} \approx \boxed{1.29}$

8 0

2 years ago