Question 1 (1 point)

Mathematics

2 answers:

Llana [10]4 years ago

6 0

Answer: The answer is 53

nasty-shy [4]4 years ago

5 0

Answer:45

Step-by-step explanation:

[9(7-3)+13]-[11-(6+9)]

[9(4)+13]-[11-(15)]

[36+13]-[4]

49-4

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-7x+4y=24, 4x-4y=0 substitution

BartSMP [9]

The first one: -7(x) + 4(y) = 24
The second one: 4(x) - 4(y) = 0

4y - 7x = 24 which is reduced to a negative
(-4y + 4x = 0 )

4(x) = 4(y)

-7(y ) + 4(y) = 24
- 3(y) = 24

3(y) = - 24

y-intercept: - 8
x-intercept: -8

The X & Y intercept is: -8, -8

8 0

3 years ago

The automobile assembly plant you manage has a Cobb-Douglas production function given by p = 10x^0.2 y^0.8

Zepler [3.9K]

Answer:

The variation needed for the daily buget to follow the increase in production for the first year is 12.38 $/year.

This value of Δy is not constant for a constant increase in production.

Step-by-step explanation:

We know that the production function is $p = 10x^{0.2} y^{0.8}$ , and in the current situation $p=1200$ and $x=130$ .

With this information we can calculate the actual budget level:

$p_0 = 10x^{0.2} y^{0.8}\\\\1200=10*130^{0.2} y^{0.8}\\\\1200=26.47*y^{0.8}\\\\y=(1200/26.47)^{1/0.8}=45.33^{1.25}=117.62$

The next year, with an increase in demand of 100 more automobiles, the production will be $p_1=1300$ .

If we calculate y for this new situation, we have:

$y_1=(\frac{p_1}{10x^{0.2}} )^{1.25}=(\frac{1300}{26.47} )^{1.25}=49.10^{1.25}=130$

The budget for the following year is 130.

The variation needed for the daily buget to follow the increase in production for the first year is 12.38 $/year.

$\Delta y=y_1-y_0=130.00-117.62=12.38$

This value of Δy is not constant for a constant increase in production.

3 0

3 years ago

Common factors of 16a and 20ab.

sweet [91]

Answer:

$\large\boxed{Common\ factors:\ 1,\ 2,\ 4,\ a,\ 2a,\ 4a}$

Step-by-step explanation:

$16a=2\cdot2\cdot2\cdot2\cdot a\\Factors:\boxed1,\ \boxed2,\ \boxed4,\ 8,\ 16,\ \boxed{a},\ \boxed{2a},\ \boxed{4a},\ 8a,\ 16a\\\\20ab=2\cdot2\cdot5\cdot a\cdot b\\Factors:\boxed1,\ \boxed2,\ \boxed4,\ 5,\ 10,\ 20,\ \boxed{a},\ \boxed{2a},\ \boxed{4a},\ 5a,\ 10a,\ 20a,\\b,\ 2b,\ 4b,\ 5b,\ 10b,\ 20b,\ 2ab,\ 4ab,\ 5ab,\ 10ab,\ 20ab\\\\Common\ factors:\ 1,\ 2,\ 4,\ a,\ 2a,\ 4a$