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2 years ago

Please help me ASAP!!

Mathematics

1 answer:

viktelen [127]2 years ago

7 0

Answer:

Step-by-step explanation:

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A salesperson makes 18% commission on every TV sold. If the salesperson sold four TV’s at $885 each, how much commission did the

Levart [38]

Answer:

They made a $531 commission.

Step-by-step explanation:

The given is:

The price of each TV is $885

The commission is 18% on every TV sold

The salesperson sold 4 TV's

We need the value of the commission

Let us find the commission of every TV, then find the total amount of all

∵ The price of each TV is $885

∵ The commission is 18 %

→ Find 18% of $885

∴ The commission on every TV = 885 × 15%

∴ The commission on every TV = 885 × 15/100

∴ The commission on every TV = $132.75

∵ The salesperson sold 4 TV

∴ The commission on them = 4 × 132.75

∴ The commission on them = $531

They made a $531 commission.

5 0

3 years ago

The work shows how to use long division to find (x2 + 3x –9) ÷ (x – 2).

Maksim231197 [3]

2x+3x -9. You can simplify this to 5x-9.

8 0

3 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

3 years ago

The hypotenuse of a right-angled triangle is 61 cm. If one of the remaining two sides is

igomit [66]

Answer:

60cm

Step-by-step explanation:

the length of other side = sqrt (square root) hypotenuse^2 - the length of thrid side^3 = sqrt 61^2 - 11^2 = sqrt 60^2 = 60

8 0

2 years ago

Find x. (More info in pic)

Studentka2010 [4]

Answer:

it would be B.

Step-by-step explanation:

if you do 4x+20=3x

then you solve x by simplifying both sides of the equation and isolate the variable.

7 0

2 years ago