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3 years ago

Find an explicit rule for the nth term of the sequence.

Mathematics

1 answer:

otez555 [7]3 years ago

4 0

This is a GS with first term -5 and common ratio = 5.
an = a1. r^(n -1)
= -5 . 5^(n - 1)

Its D

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if the parent function is square root of x, describe the translation of this function square root x+7

kotykmax [81]

The function $y=\sqrt{x}$ sits with its starting point at the origin. The function $y=\sqrt{(x-h)}+k$ is translated h units to the left or right and k units up or down. Since our radicand, the value under the square root sign, is x+7, putting that into our standard form it would be x-(-7) because minus a negative is a positive. So we move the parent graph to the left (-7 moves to the left) 7 units. There is no number k so our function is not moving up or down. Only to the left 7.

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3 years ago

Wite the equation of the linear relationship in slope-intercept form. Help me please:))

Digiron [165]

Answer:

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Step-by-step explanation:

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3 years ago

2 (h-8)-h = h-16 what is the solution

Svetach [21]

Let's solve your equation step-by-step.2(h−8)−h=h−16Step 1: Simplify both sides of the equation.2(h−8)−h=h−16Simplify: (Show steps)h−16=h−16Step 2: Subtract h from both sides.h−16−h=h−16−h−16=−16
Step 3: Add 16 to both sides.−16+16=−16+160=0Answer:All real numbers are solutions.

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3 years ago

New York City has a population of 8.55 million and 300 square mile. Manhattan has 1.64 million and 23 square miles. How much gre

harkovskaia [24]

Given :

New York City has a population of 8.55 million and 300 square mile area .

Manhattan has 1.64 million and 23 square miles area .

To Find :

How much greater is the population density of Manhattan than New York City.

Solution :

Population density of New York City :

$D_{NY}=\dfrac{8.55}{300}=0.029 \ million/mile^2\\\\D_{M}=\dfrac{1.64}{23}=0.071 \ million/mile^2$

Now ,

$D_{M}-D_{NY}\\\\(0.071-0.029)=0.042\ million/miles^2$

Therefore , population density of Manhattan is $0.042\ million/miles^2$ greater than New York City .

Hence , this is the required solution .

4 0

3 years ago

A veterinarian weighed a sample of 666 puppies. Here are each of their weights (in kilograms): 1, 2, 7, 7, 10, 151,2,7,7,10,151,

rjkz [21]

Answer:

The standard deviation of the weight, σ, is approximately 4.73 kg

Step-by-step explanation:

The weights of the six puppies are;

$x_i$ = 1, 2, 7, 7, 10, and 15

The number of puppies, n = 6

Therefore, we have;

The mean = μ = ∑x/n = (1 + 2 + 7 + 7 + 10 + 15)/6 = 7

The standard deviation, σ, is given as follows;

$\sigma =\sqrt{\dfrac{\sum \left (x_i-\mu \right )^{2} }{n}}$

By substituting, we have;

$\sigma =\sqrt{\dfrac{ \left (1-7 \right )^{2} + \left (2-7 \right )^{2} + \left (7-7 \right )^{2} + \left (7-7 \right )^{2} + \left (10-7 \right )^{2} + \left (15-7 \right )^{2} }{6}}$

Simplifying gives;

$\sigma =\sqrt{\dfrac{ \left (-6 \right )^{2} + \left (-5 \right )^{2} + \left (0 \right )^{2} + \left (0 \right )^{2} + \left (3 \right )^{2} + \left (8 \right )^{2} }{6}} =\sqrt{\dfrac{ 36 + 25 + 0 +0+ 9 + 64 }{6}}$

$\sigma =\sqrt{\dfrac{ 134 }{6}} = \sqrt{\dfrac{67}{3} } \approx 4.72582$

The standard deviation of the weight, σ ≈ 4.73 kg to two decimal places.

4 0

3 years ago

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