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Karo-lina-s [1.5K]

3 years ago

13

Amersham station is the highest station in the london tube at 147m above sea level. Waterloo statin is the lowest startion at 21

m below sea level what is the height difference

1 answer:

Serjik [45]3 years ago

4 0

168m - it is positive 147 - (-21), so the difference is 168m

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muminat

Answer:

100 ounce

7÷140=20

20×5=100

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2 years ago

Item 22 Question 1 You have a coupon for 10% off the sale price of a surfboard. Which is the better buy? Explain your reasoning.

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The first question 10 and the next one is 40 the next one is 30

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2 years ago

The function shown below was created to track the different intervals of speed that an automobile travels over a period of 28 se

Ksenya-84 [330]

Answer:

1. The car is slowing down at a rate of 2.5mph/s

2. The greatest acceleration is 10 mph/s.

3. In the interval 4s to 16s the speed remains constant and has magnitude 25 mph.

Step-by-step explanation:

1. The deceleration of the car is from 16 seconds to 24 seconds is the slope $m$ of the graph from 16 to 24:

$m=\dfrac{\Delta speed }{\Delta time } = \dfrac{5-25}{24-16} =-2.5mph/s$

the negative sign indicates that it is deceleration.

2. The automobile experiences the greatest change in speed when the slope is greatest because that is when acceleration/deceleration is greatest.

From the graph we see that the greatest slope of the graph is between 28 and 24 seconds. The acceleration the interval is the slope $m$ :

$m= \dfrac{45-5}{28-24}= 10mph/s$

3. The automobile experiences no acceleration in the interval 4 s to 16 s—that's the graph is flat.

The speed of the automobile in that interval, as we see from the graph, is 25 mph.

7 0

3 years ago

Read 2 more answers

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Aleks04 [339]

110? It's a supplementary angle.

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2 years ago

Read 2 more answers

John, Sally, and Natalie would all like to save some money. John decides that it

brilliants [131]

Answer:

Part 1) John’s situation is modeled by a linear equation (see the explanation)

Part 2) $y=100x+300$

Part 3) $\$12,300$

Part 4) $\$2,700$

Part 5) Is a exponential growth function

Part 6) $A=6,000(1.07)^{t}$

Part 7) $\$11,802.91$

Part 8) $\$6,869.40$

Part 9) Is a exponential growth function

Part 10) $A=5,000(e)^{0.10t}$ or $A=5,000(1.1052)^{t}$

Part 11) $\$13,591.41$

Part 12) $\$6,107.01$

Part 13) Natalie has the most money after 10 years

Part 14) Sally has the most money after 2 years

Step-by-step explanation:

Part 1) What type of equation models John’s situation?

Let

y ----> the total money saved in a jar

x ---> the time in months

The linear equation in slope intercept form

y=mx+b

The slope is equal to

$m=\$100\ per\ month$

The y-intercept or initial value is

$b=\$300$

so

$y=100x+300$

therefore

John’s situation is modeled by a linear equation

Part 2) Write the model equation for John’s situation

see part 1)

Part 3) How much money will John have after 10 years?

Remember that

1 year is equal to 12 months

so

$10\ years=10(12)=120 months$

For x=120 months

substitute in the linear equation

$y=100(120)+300=\$12,300$

Part 4) How much money will John have after 2 years?

Remember that

1 year is equal to 12 months

so

$2\ years=2(12)=24\ months$

For x=24 months

substitute in the linear equation

$y=100(24)+300=\$2,700$

Part 5) What type of exponential model is Sally’s situation?

we know that

The compound interest formula is equal to

$A=P(1+\frac{r}{n})^{nt}$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

n is the number of times interest is compounded per year

in this problem we have

$P=\$6,000\\ r=7\%=0.07\\n=1$

substitute in the formula above

$A=6,000(1+\frac{0.07}{1})^{1*t}\\ A=6,000(1.07)^{t}$

therefore

Is a exponential growth function

Part 6) Write the model equation for Sally’s situation

see the Part 5)

Part 7) How much money will Sally have after 10 years?

For t=10 years

substitute the value of t in the exponential growth function

$A=6,000(1.07)^{10}=\$11,802.91$

Part 8) How much money will Sally have after 2 years?

For t=2 years

substitute the value of t in the exponential growth function

$A=6,000(1.07)^{2}=\$6,869.40$

Part 9) What type of exponential model is Natalie’s situation?

we know that

The formula to calculate continuously compounded interest is equal to

$A=P(e)^{rt}$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

e is the mathematical constant number

we have

$P=\$5,000\\r=10\%=0.10$

substitute in the formula above

$A=5,000(e)^{0.10t}$

Applying property of exponents

$A=5,000(1.1052)^{t}$

therefore

Is a exponential growth function

Part 10) Write the model equation for Natalie’s situation

$A=5,000(e)^{0.10t}$ or $A=5,000(1.1052)^{t}$

see Part 9)

Part 11) How much money will Natalie have after 10 years?

For t=10 years

substitute

$A=5,000(e)^{0.10*10}=\$13,591.41$

Part 12) How much money will Natalie have after 2 years?

For t=2 years

substitute

$A=5,000(e)^{0.10*2}=\$6,107.01$

Part 13) Who will have the most money after 10 years?

Compare the final investment after 10 years of John, Sally, and Natalie

Natalie has the most money after 10 years

Part 14) Who will have the most money after 2 years?

Compare the final investment after 2 years of John, Sally, and Natalie

Sally has the most money after 2 years

3 0

3 years ago