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marta [7]
3 years ago
11

1. Pilar and her sister Marisol live near the beach. Pilar has 80 seashells in her collection when Marisol started collecting se

ashells. Marisol started with zero seashells in her collection.
• Pilar adds 10 seashells to her collection each week for x weeks.
• Marisol adds 20 seashells to her collection each week for x weeks.
(a) Write an inequality to represent the situation when the number of seashells in Marisol's collection is greater than the number of seashells in Pilar's collection.
(b) Solve the inequality in part (a) for the x. Show your work and explain your answer.
Mathematics
1 answer:
enot [183]3 years ago
3 0

Answer:

a. 20x > 10x + 80

b.  •20x > 10x + 80

• Subtract 10x from both sides: 20x -10x > 10x + 80 - 10x

• Simplify: 10x > 80

• Divide both sides by 10:   10x/10 > 80/10

• Simplify: x > 8

Step-by-step explanation:

10x + 80 = number of seashells Pilar has.

20x = number of seashells Marisol has.  

answer for a:  20x > 10x + 80

answer for b: 20x > 10x + 80

• Subtract 10x from both sides: 20x -10x > 10x + 80 - 10x

• Simplify: 10x > 80

• Divide both sides by 10:   10x/10 > 80/10

• Simplify: x > 8

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Evaluate S5 for 300 + 150 + 75 + … and select the correct answer below. 18.75 93.75 581.25 145.3125
Savatey [412]

we have that

300 + 150 + 75 +...

Let

a1=300\\ a2=150\\ a3=75

we know that

\frac{a2}{a1} =\frac{150}{300} \\\\ \frac{a2}{a1}=0.5 \\ \\ a2=a1*0.50

\frac{a3}{a2} =\frac{75}{150} \\\\ \frac{a3}{a2}=0.5 \\ \\ a3=a2*0.50

so

a(n+1)=an*0.50

Is a geometric sequence

Find the value of a4

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a(4)=75*0.50

a(4)=37.5

Find the value of a5

a(5)=a4*0.50

a(5)=37.5*0.50

a(5)=18.75

Find S5

S5=a1+a2+a3+a4+a5\\ S5=300+150+75+37.5+18.75\\ S5=581.25

therefore

the answer is

581.25

Alternative Method

Applying the formula

S_n=\frac{a_1 (1-r^n)}{1-r} \\\\a_1=300 \\ r=\frac{1}{2}\\\\ S_5=\frac{300(1-(\frac{1}{2})^5)}{1-\frac{1}{2}}\\\\=\frac{300(1-\frac{1}{32})}{\frac{1}{2}}\\\\=\frac{300 \times \frac{31}{32}}{\frac{1}{2}}\\\\=\frac{75 \times \frac{31}{8}}{\frac{1}{2}}\\\\=\frac{\frac{2325}{8}}{\frac{1}{2}}\\\\=\frac{2325}{8} \times 2\\\\=\frac{2325}{4}\\\\=581 \frac{1}{4}\\\\=581.25

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581.25

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Answer:

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n = number of compoundings per year. In our case 2 (semiannually)

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