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2 years ago

Jikihkhuhjknbnmjk ijklnmjb ?

Mathematics

1 answer:

Softa [21]2 years ago

7 0

Dhkwkgk k sad h e nah dawg he’s as Jewish

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A 5000 seat theater has tickets for sale at 26 and 40. How many tickets should be sold at each price for a sellout performance t

slavikrds [6]

Answer:

The number of tickets for sale at $26 should be 3300

The number of tickets for sale at $40 should be 1700

Step-by-step explanation:

Use 2 equations to represent the modifiers within the problem:

$5000 = a + b \\ 153800 = 26a + 40b$

Now you want to find the point at which the variables are changed to make both equations correct, this can be done by graphing and finding the intersection of both lines.

$5000 = 3300 + 1700 \\ 153800 = 26(3300) + 40(1700)$

5 0

2 years ago

Answer me pleaseeee..... prime numbers between 1 to 10 ?

My name is Ann [436]

Answer:

2, 3, 5, 7

Step-by-step explanation:

prime numbers between 1 to 10 are 2,3,5,7. it's very easy. 1 is not prime number because it behave as both prime and composite.

8 0

1 year ago

Are the graphs of each pair of equations parallel, perpendicular, or neither? x=4 y=4

Vlad1618 [11]

Both graphs x =4 and y =4 will be perpendicular to each other.

The graph of x = 4 is a vertical line down the Graph scope
The graph of y =4 is a horizontal line side the Graph scope.

Both the Lines will meet exactly at ( 4, 4 ) perpendicularly at each other both of them intersecting at 90 degrees.

Vertical and Horizontal lines are perpendicular to each other (90 Degrees).

Know more about Perpendicular Lines: brainly.com/question/1202004

5 0

3 months ago

Read 2 more answers

Ty received test grades of 82%, 74%, 76%, 76%, and 74%.

Nikitich [7]

Answer:

a) To earn C he needs at least 68 and less than 98

b) No, he couldn't make it.

Step-by-step explanation:

I. His avg value is (382 + x)/6

I pick a random number and that gave

(68 => to reach 75) and (98 => to reach 80)

So that makes he needs at least 68 and less than 98

II. The reason why he cannot make it is

if 100 is 100%

= (100 + 382)/6

= 80.33 That's impossible to reach 90

Hope that help :)

7 0

2 years ago

For the given term, find the binomial raised to the power, whose expansion it came from: 15(5)^2 (-1/2 x) ^4

Elina [12.6K]

Answer:

C. $(5-\frac{1}{2})^6$

Step-by-step explanation:

Given

$15(5)^2(-\frac{1}{2})^4$

Required

Determine which binomial expansion it came from

The first step is to add the powers of he expression in brackets;

$Sum = 2 + 4$

$Sum = 6$

Each term of a binomial expansion are always of the form:

$(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......$

Where n = the sum above

$n = 6$

Compare $15(5)^2(-\frac{1}{2})^4$ to the above general form of binomial expansion

$(a+b)^n = ......+15(5)^2(-\frac{1}{2})^4+.......$

Substitute 6 for n

$(a+b)^6 = ......+15(5)^2(-\frac{1}{2})^4+.......$

[Next is to solve for a and b]

From the above expression, the power of (5) is 2

Express 2 as 6 - 4

$(a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......$

By direct comparison of

$(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......$

and

$(a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......$

We have;

$^nC_ra^{n-r}b^r= 15(5)^{6-4}(-\frac{1}{2})^4$

Further comparison gives

$^nC_r = 15$

$a^{n-r} =(5)^{6-4}$

$b^r= (-\frac{1}{2})^4$

[Solving for a]

By direct comparison of $a^{n-r} =(5)^{6-4}$

$a = 5$

$n = 6$

$r = 4$

[Solving for b]

By direct comparison of $b^r= (-\frac{1}{2})^4$

$r = 4$

$b = \frac{-1}{2}$

Substitute values for a, b, n and r in

$(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......$

$(5+\frac{-1}{2})^6 = ......+ ^6C_4(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ ^6C_4(5)^{6-4}(\frac{-1}{2})^4+.......$

Solve for $^6C_4$

$(5-\frac{1}{2})^6 = ......+ \frac{6!}{(6-4)!4!)}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{6!}{2!!4!}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{6*5*4!}{2*1*!4!}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{6*5}{2*1}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{30}{2}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+15*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+15(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+15(5)^2(\frac{-1}{2})^4+.......$

Check the list of options for the expression on the left hand side

The correct answer is $(5-\frac{1}{2})^6$

3 0

2 years ago