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3 years ago

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Mr. and Mrs. Rodriguez are taking their children to Colonial Williamsburg. A 1-day admission pass is $33 for an adult and $16.50

for a child. If they pay a total of $115.50 admission for themselves and their children, how many children do they have?

2 answers:

EleoNora [17]3 years ago

8 0

The answer is 3 kids. hope i helped!!!!!!!

quester [9]3 years ago

7 0

They have 3 kids hope that helped

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Find the vertical and horizontal asympotote for (×-2) (×+3)/ (2×+2)

nekit [7.7K]

$\bf \cfrac{(x-2)(x+3)}{2x+2}\implies \cfrac{x^2+x-6}{2x+2}~~ \begin{array}{llll} \leftarrow \textit{2nd degree polynomial}\\ \leftarrow \textit{1st degree polynomial} \end{array} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{vertical asymptote}}{2x+2=0}\implies 2x=-2\implies x=-\cfrac{2}{2}\implies x=-1$

when the degree of the numerator is greater than the denominator's, then it has no horizontal asymptotes.

quick note:

when the degree of the numerator is 1 higher than the degree of the denominator, then it has an slant-asymptote, so this one has a slant-asymptote.

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3 years ago

Anyone here who can help me with this?

Mandarinka [93]

Answer:

D. 16 years old

Step-by-step explanation:

<u>Step 1:</u> Let T be Tien's age and J as Jordan's age (today),

$T=\frac{1}{4}J$

<u>Step 2:</u> Let T be Tien's age and J as Jordan's age (in 2 years),

$T+2=\frac{1}{3} (J+2)$

$T=\frac{1}{3}(J+2)-2$

<u>Step 3:</u> As their age differences will always be similar we can have the two equations above equal to find Jordan's age,

$\frac{1}{4} J=\frac{1}{3}(J+2)-2\\\frac{1}{4}J-\frac{1}{3}J=\frac{2}{3} -2 \\-\frac{1}{12}J= -\frac{4}{3} \\\\J=16$

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3 years ago

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Elodia [21]

2.5/0.2=12.5
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BartSMP [9]

Answer:

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Volgvan

Answer:

II and II only.

Step-by-step explanation:

The first series converges because the common ratio r is 0.8 (<1).

II diverges as |r| > 1.

III is an alternating series with r = -4 - that is |r| = 4 so it diverges.

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3 years ago