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2 years ago

5

Can anyone please help me with this?

1 answer:

marin [14]2 years ago

4 0

Answer:

I believe that It would be b

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What I need to do with this problems

alina1380 [7]

Answer:

L x W x H =Area

Step-by-step explanation:

8 0

3 years ago

A person runs 1/8 mile in 1/64 hour.<br> The person's speed is (BLANK) miles per hour.

snow_tiger [21]

Answer: The person's speed is 8 miles per hour.

Step-by-step explanation:

To solve we want to convert the $\frac{1}{8}$ and $\frac{1}{64}$ proportionally for the number of miles per hour. We know that multiplying a fraction by its reciprocal will equal 1. So let's do that for the value of the hour.

$\frac{1}{64}$ × $\frac{64}{1} =\frac{64}{64} =1$

Since we multiplied that to the number of hours we must also do that for the number of miles.

$\frac{1}{8}$ × $\frac{64}{1} =\frac{64}{8} =8$

7 0

1 year 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

2 years ago

hi I need help with this question

motikmotik

Answer:

Use your calculator to find the value of:

Write all the figures on your calculator's display for th

5 0

2 years ago

Refer to the question below:

irakobra [83]

∛(2x + 5) - 3 = 0
Add 3 to both sides
∛(2x + 5) = 3
Cube both sides
2x + 5 = 27
Subtract 5 from both sides
2x = 22
Divide both sides by 2
x=11
There's your answer. Have an awesome day! :)

7 0

3 years ago