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geniusboy [140]

3 years ago

6

Stan is on a bike ride .His gps indicates that he has traveled exactly 18 miles and he is 32% complete with his total ride.How m

any miles long is stans ride

1 answer:

egoroff_w [7]3 years ago

3 0

32% of his total ride is 18 miles

0.32x = 18
x = 18 / 0.32
x = 56.25 or 56 1/4 miles <== total distance

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X varies directly with<br> y. If x = 60 when<br> y = 75. find x when<br> y = 32.

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Answer:

x=17

Step-by-step explanation:

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2 years ago

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

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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 />

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2 years ago

Alan is comparing the cost of a fresh lobster dinner at two different restaurants. The first

Pepsi [2]

Good evening ,

Answer:

The weight : 29 pound

Alan would pay for his dinner : $130

Step-by-step explanation:

Let x represent the weight of the lobster

the cost of the dinner at restaurant 1 is 4x + 14

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⇔ x = 43 - 14

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In this case Alan would pay the same amount for his dinner at any restaurant:

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= 3×(29)+43 = 130.

:)

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2 years ago

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Answer:

i think the answe is A and B hope this helps

Step-by-step explanation:

Have a nice day!

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1 year ago

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Given the circle below secant kjI and tangent hi find the length of hi round to the nearest tenth if necessary.

garik1379 [7]

The length of the segment HI in the figure is 32.9

<h3>How to determine the length HI?</h3>

To do this, we make use of the following secant-tangent equation:

HI² = KI * JI

From the figure, we have:

KI = 21 + 24 = 45

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So, we have:

HI² = 45 * 24

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1 year ago