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Ksju [112]
2 years ago
9

Select the equation that represents the line that passes through (1,7) and is parallel to y = 5x + 4

Mathematics
2 answers:
saveliy_v [14]2 years ago
5 0

Answer:

The equation that represents the line that passes through (1,7) and is parallel to y = 5x + 4 will be:

  • y=5x+2

Step-by-step explanation:

Given the equation

y = 5x + 4

Here the slope is:

m=5

as comparing with the y=mx+b, which is the slop-intercept form where m is the slope and b is the y-intercept.

As we know the point-slope form is

y-y_1=m\:\left(x-x_1\right)

y-7=5\left(x-1\right)

Add 7 to both sides

y-7+7=5\left(x-1\right)+7

y=5x+2

Therefore, the equation that represents the line that passes through (1,7) and is parallel to y = 5x + 4 will be:

  • y=5x+2
VladimirAG [237]2 years ago
3 0

Answer:

a

Step-by-step explanation:

it has to  have to same slope as the first equation so you have y=5x+b then what you do is plug in the points and solve for b. which gives you 2. so your final equation is y=5x+2

7=5+b

-5  -5

2=b

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3 years ago
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Find the area of the composite figure
-BARSIC- [3]

Answer:

48 ft^2

Step-by-step explanation:

5.5 times 6 equals 33 (larger square part)

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3 years ago
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

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photoshop1234 [79]

Answer:

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Step-by-step explanation:

You always work from the parenthesis first because of order of operation. You have 12( because 2 times 6) then mulitply 12 times 10. You get 120 then you multiply it but itself 3 times. 120 x 120 x 120= 1728000

6 0
3 years ago
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WINSTONCH [101]

Answer:

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Step-by-step explanation:

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f(3) means u have to put the value of 3 in the given equation so it will be

= 7-2×3

= 7-6

= 1

and now you have to add 6 to it so it will be 7.

8 0
3 years ago
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