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2 years ago

10

Bradley owns 2 surfboards and 4 wet suits. If he takes one surfboard and one wet suit to the beach, how many different combinati

ons can he choose?

1 answer:

photoshop1234 [79]2 years ago

7 0

Answer:

The answer is eight

Step-by-step explanation:

4*2=8

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Given: - 4x/7 > 10. Choose the solution set. {x | x < - } {x | x > - } {x | x > - } {x | x < - }

AleksAgata [21]

Answer:

{x | x < - 35 / 2}

Step-by-step explanation:

-4x / 7 > 10

Multiply both sides by 7

-4x /7 * 7 > 10 * 7

-4x > 70

Divide both sides by - 4

-4x / - 4 > 70 / - 4

x < - 35 / 2 (inequality sign changes)

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2 years ago

Which of the following functions is graphed below?

11Alexandr11 [23.1K]

We can't tell from the graph whether the left point of each step is included, or the right point. If the left, this is a graph of
y = 7 + floor(x)

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3 years ago

Read 2 more answers

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

A small box measures 7 in. by 10 in. by <br> 3/4 in. high. Find the volume of the box.

vitfil [10]

Answer:

Volume = 52.5

Step-by-step explanation:

volume of a rectangular 3d shape can be calculated by length*width*height. In this case:

7*10*0.75 = 52.5

8 0

3 years ago

Please help me. these problems<br>

jeyben [28]

Answer:

1st problem:

Converges to 6

2nd problem:

Converges to 504

Step-by-step explanation:

You are comparing to $\sum_{k=1}^{\infty} a_1(r)^{k-1}$

You want the ratio r to be between -1 and 1.

Both of these problem are so that means they both have a sum and the series converges to that sum.

The formula for computing a geometric series in our form is $\frac{a_1}{1-r}$ where $a_1$ is the first term.

The first term of your first series is 3 so your answer will be given by:

$\frac{a_1}{1-r}=\frac{3}{1-\frac{1}{2}}=\frac{3}{\frac{1}{2}=6$

The second series has r=1/6 and a_1=420 giving me:

$\frac{420}{1-\frac{1}{6}}=\frac{420}{\frac{5}{6}}=420(\frac{6}{5})=504$ .

3 0

3 years ago