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2 years ago

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Kendra swims 126 miles in 3 days, what is the constant rate that Kendra swims per day? a. 40 miles per day c. 42 miles per day b

. 41 miles per day d. 43 miles per day

1 answer:

Leni [432]2 years ago

5 0

Answer:

c. 42 miles per day

Step-by-step explanation:

From the information given, you can use a rule of three to calculate the constant rate that Kendra swims per day given that she swims 126 miles in 3 days:

3 days → 126 miles

1 day → x

x=(1*126)/3=42 miles

According to this, the answer is that Kendra swims 42 miles per day.

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Find g(a-2)+3g(2a) if g(x)=x²-5x+8

sertanlavr [38]

Answer:

13a² - 39a + 46

Step-by-step explanation:

To find g(a-2)+3g(2a), find each part using the function g(x)=x²-5x+8.

g(a-2) = (a-2)²-5(a-2)+8 = a² - 4a + 4-5a + 10+8 = a² - 9a + 22

3g(2a) = 3{(2a)²-5(2a)+8} = 3{ 4a² - 10a + 8} = 12a² - 30a + 24

Combine the values to find g(a-2)+3g(2a).

g(a-2)+3g(2a) = (a² - 9a + 22) + (12a² - 30a + 24) = 13a² - 39a + 46

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3 years ago

I don’t understand what to do

Natasha2012 [34]

Answer: 14cm

Step-by-step explanation:

The diameter of a circle is twice the radius.

Therefore, the diameter = 2*7 cm = 14cm

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2 years ago

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Write an equation of a circle that is centered at (1,-4) with a radius of 10.

alexdok [17]

Answer:

$(x-1)^2 + (y+4)^2 = 100\\\\$

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

Work Shown:

$(x-h)^2 + (y-k)^2 = r^2\\\\(x-1)^2 + (y-(-4))^2 = 10^2\\\\(x-1)^2 + (y+4)^2 = 100\\\\$

You could expand terms out and simplify, but I think it's more handy to leave it in this form so you can easily spot the center and radius from a glance.

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2 years ago

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Match each series with the equivalent series written in sigma notation

PIT_PIT [208]

Answer:

$3 + 12 + 48 + 192 + 768 = \sum\limits^4_{n=0} 3 * 4^n$

$4 + 32 + 256 + 2048 + 16384 = \sum\limits^4_{n=0} 4 * 8^n$

$2 + 6 + 18 + 54 + 162 = \sum\limits^4_{n=0} 2* 3^n$

$3 + 15 + 75 + 375 + 1875 = \sum\limits^4_{n=0} 3* 5^n$

Step-by-step explanation:

Given

See attachment for complete question

Required

Match equivalent expressions

Solving (a):

$3 + 12 + 48 + 192 + 768$

The expression can be written as:

$3 \to 3*4^{0$ --- 0

$12 \to 3 * 4^{1$ ---- 1

$48 \to 3 * 4^{2$ --- 2

$192 \to 3 * 4^{3$ ---- 3

$768 \to 3 * 4^{4$ ---- 4

For the nth term, the expression is:

$Term = 3 * 4^{n$ ---- n

So, the summation is:

$3 + 12 + 48 + 192 + 768 = \sum\limits^4_{n=0} 3 * 4^n$

Solving (b):

$4 + 32 + 256 + 2048 + 16384$

The expression can be written as:

$4 \to 4 * 8^0$ --- 0

$32 \to 4 * 8^1$ ---- 1

$256 \to 4 * 8^2$ --- 2

$2048 \to 4 * 8^3$ ---- 3

$16384 \to 4 * 8^4$ ---- 4

For the nth term, the expression is:

$Term \to 4 * 8^n$ ---- n

So, the summation is:

$4 + 32 + 256 + 2048 + 16384 = \sum\limits^4_{n=0} 4 * 8^n$

Solving (c):

$2 + 6 + 18 + 54 + 162$

The expression can be written as:

$2 \to 2 * 3^0$ --- 0

$6 \to 2 * 3^1$ ---- 1

$18 \to 2 * 3^2$ --- 2

$54 \to 2 * 3^3$ ---- 3

$162 \to 2 * 3^4$ ---- 4

For the nth term, the expression is:

$Term \to 2 * 3^n$ ---- n

So, the summation is:

$2 + 6 + 18 + 54 + 162 = \sum\limits^4_{n=0} 2* 3^n$

Solving (d):

$3 + 15 + 75 + 375 + 1875$

The expression can be written as:

$3 \to 3 * 5^0$ --- 0

$15 \to 3 * 5^1$ ---- 1

$75 \to 3 * 5^2$ --- 2

$375 \to 3 * 5^3$ ---- 3

$1875 \to 3 * 5^4$ ---- 4

For the nth term, the expression is:

$Term \to 3 * 5^n$ ---- n

So, the summation is:

$3 + 15 + 75 + 375 + 1875 = \sum\limits^4_{n=0} 3* 5^n$

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Answer:

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Step-by-step explanation:

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3 years ago

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