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3 years ago

5

Which is the better buy? 12 bars of soap for $10.00 or 5 bars of soap for $4.00

2 answers:

Butoxors [25]3 years ago

6 0

It is cheaper to buy 2 5packs because your total will be $8 not $10 so you end up saving $2 a dollar per pack

likoan [24]3 years ago

3 0

5 Bars of soap for $4.00 Why? Because you if you buy two it cost $8.00 or buy three pack and you'll get 15 for $12.00

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Diego was asked to plot these points: (-50, 0), (150, 100), (200, -100), (350, 50), (-250, 0). What interval could he use for ea

ivolga24 [154]

Answer:

Interval of 50 on both axis

Step-by-step explanation:

Given

$(x_1,y_1) = (-50,0)$

$(x_2,y_2) = (150,100)$

$(x_3,y_3) = (200,-100)$

$(x_4,y_4) = (350,50)$

$(x_5,y_5) = (-250,0)$

There are several ways to do this, but I will use the observation method, since the dataset is small.

Considering the x-coordinates

$x = \{-50,150,200,350,-250}$

Each element of the data set is a multiple of 50.

Hence, an interval of 50 can be used on the x-axis

Considering the y-coordinates

$y = \{0,100,-100,50,0\}$

Each element of the data set is a multiple of 50.

Hence, an interval of 50 can be used on the y-axis

<em>So, an interval of 50 can be used on both axes</em>

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3 years ago

Please help I’ll mark you as brainliest if correct

Dominik [7]

Answer:

25%

Step-by-step explanation:

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3 years ago

Help me please and explain so i can understand

Finger [1]

The answer is B
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3 years ago

A television costs $450. During a special sale, it’s marked 1\3 off. What’s the sales price of the tv?

Darina [25.2K]

Answer:

150

Step-by-step explanation:

4 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:

<em>C.</em> $(5-\frac{1}{2})^6$

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]

<em>From the above expression, the power of (5) is 2</em>

<em>Express 2 as 6 - 4</em>

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

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 $^6C_4$

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

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

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

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

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

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

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

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

<em>Check the list of options for the expression on the left hand side</em>

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

3 0

3 years ago