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Ivan
3 years ago
14

Solve the proportion 6/y=9/24

Mathematics
2 answers:
anzhelika [568]3 years ago
7 0

Answer:

It would be 16!

Step-by-step explanation:

Step 1 : 6 times 24 is 144

Step 2 : 144/9

Answer 16!

Leno4ka [110]3 years ago
6 0

Answer:

ur dad and. ur mom plus ur dad mama ififjfifnfbfndjfkjfif

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<h3>Answer: Choice B</h3>

=============================================

Explanation:

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Now let's turn to choice D. If we multiply both sides of the first inequality by -1, then we go from -2x-5y \le 14 to 2x+5y \ge -14. Note the inequality sign flips. This always happens when we multiply both sides by any negative number. The inequality 2x+5y \ge -14 implies that the shaded region will be above the boundary line, but this contradicts the drawing which shows the shaded region is below the diagonal boundary line. We can rule out choice D because of this. Choice A can be ruled out for similar reasoning.

You should find that only choice B is left. The diagonal line is 2x+5y = 14, and we shade below this boundary line, as well as shading to the right of the y axis (to indicate all values have positive x coordinates). Values on the boundary count as solution points as well.

3 0
3 years ago
Please help with this question!!!
kondor19780726 [428]
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7 0
3 years ago
4. Benson sells vacuum cleaners on a commission. He earns $75 for each vacuum cleaner he sells.
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Read 2 more answers
Geometric Sequence S = 1.0011892 + ... + 1.0012 + 1.001 + 1
leva [86]

Answer:

<em />S_{1893} =5632.98<em />

<em />

Step-by-step explanation:

The correct form of the question is:

S = 1.001^{1892} + ... + 1.001^2 + 1.001 + 1

Required

Solve for Sum of the sequence

The above sequence represents sum of Geometric Sequence and will be solved using:

S_n = \frac{a(1 - r^n)}{1 - r}

But first, we need to get the number of terms in the sequence using:

T_n = ar^{n-1}

Where

a = First\ Term

a = 1.001^{1892}

r = common\ ratio

r = \frac{1}{1.001}

T_n = Last\ Term

T_n = 1

So, we have:

T_n = ar^{n-1}

1 = 1.001^{1892} * (\frac{1}{1.001})^{n-1}

Apply law of indices:

1 = 1.001^{1892} * (1.001^{-1})^{n-1}

1 = 1.001^{1892} * (1.001)^{-n+1}

Apply law of indices:

1 = 1.001^{1892-n+1}

1 = 1.001^{1892+1-n}

1 = 1.001^{1893-n}

Represent 1 as 1.001^0

1.001^0 = 1.001^{1893-n}

They have the same base:

So, we have

0 = 1893-n

Solve for n

n = 1893

So, there are 1893 terms in the sequence given.

Solving further:

S_n = \frac{a(1 - r^n)}{1 - r}

Where

a = 1.001^{1892}

r = \frac{1}{1.001}

n = 1893

So, we have:

S_{1893} =\frac{1.001^{1892} *(1 -\frac{1}{1.001}^{1893})}{1 -\frac{1}{1.001} }

S_{1893} =\frac{1.001^{1892} *(1 -\frac{1}{1.001}^{1893})}{\frac{1.001 -1}{1.001} }

S_{1893} =\frac{1.001^{1892} *(1 -\frac{1}{1.001}^{1893})}{\frac{0.001}{1.001} }

S_{1893} =\frac{1.001^{1892} *(1 -\frac{1}{1.001^{1893}})}{\frac{0.001}{1.001} }

Simplify the numerator

S_{1893} =\frac{1.001^{1892}  -\frac{1.001^{1892}}{1.001^{1893}}}{\frac{0.001}{1.001} }

S_{1893} =\frac{1.001^{1892}  -1.001^{1892-1893}}{\frac{0.001}{1.001} }

S_{1893} =\frac{1.001^{1892}  -1.001^{-1}}{\frac{0.001}{1.001} }

S_{1893} =(1.001^{1892}  -1.001^{-1})/({\frac{0.001}{1.001} })

S_{1893} =(1.001^{1892}  -1.001^{-1})*{\frac{1.001}{0.001}}

S_{1893} =\frac{(1.001^{1892}  -1.001^{-1}) * 1.001}{0.001}

Open Bracket

S_{1893} =\frac{1.001^{1892}* 1.001  -1.001^{-1}* 1.001 }{0.001}

S_{1893} =\frac{1.001^{1892+1}  -1.001^{-1+1}}{0.001}

S_{1893} =\frac{1.001^{1893}  -1.001^{0}}{0.001}

S_{1893} =\frac{1.001^{1893}  -1}{0.001}

S_{1893} =5632.97970294

Hence, the sum of the sequence is:

<em />S_{1893} =5632.98<em> ----- approximated</em>

4 0
3 years ago
javier bought 48 sports cads at a yard sale. of the cards, 3 eighths were baseball cards. how many cards were baseballcards
Sergio039 [100]
Ahemmmm   what is 3/8 of 48?  well, is just their product.

\bf \cfrac{3}{8}\cdot 48\implies \cfrac{144}{8}\implies 18
3 0
3 years ago
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