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fomenos
3 years ago
13

What is an equivalent equation for 5 potato chips in 1/4 of a min​

Mathematics
1 answer:
Mashutka [201]3 years ago
8 0
The equivalent is 59
You might be interested in
The commission for selling a house is$11825, which is 5.5% of the selling price. What is the selling price of the house?
Rzqust [24]
In this case, the best method starts with finding one percent of the selling price.  To do this, divide the commission ($11825) by 5.5.  This gets you 1%, which is $2150.  Next, multiply this number by 100, turning 1% into 100%.  Your final answer is $215,000.
8 0
3 years ago
Lines that are tangent to a circle are
jonny [76]

Step-by-step explanation:

Lines that are tangent to a circle are perpendicular to the radius of the circle.

Lines that are tangent to a circle are congruent

5 0
4 years ago
John, sally, Natalie would all like to save some money. John decides that it would be best to save money in a jar in his closet
stiv31 [10]

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)

y=100x+300

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

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

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
4 years ago
Read 2 more answers
NEED ANSWERED ASAP WILL REWARD BRAINLIEST
Luda [366]

Answer:

The sum of the first 100 terms is 60400

Step-by-step explanation:

* Lets revise the arithmetic sequence

- There is a constant difference between each two consecutive

  numbers

- Ex:

# 2  ,  5  ,  8  ,  11  ,  ……………………….

# 5  ,  10  ,  15  ,  20  ,  …………………………

# 12  ,  10  ,  8  ,  6  ,  ……………………………

* General term (nth term) of an Arithmetic sequence:

- U1 = a  ,  U2  = a + d  ,  U3  = a + 2d  ,  U4 = a + 3d  ,  U5 = a + 4d

- Un = a + (n – 1)d, where a is the first term , d is the difference

 between each two consecutive terms n is the position of the

 number

- The sum of first n terms of an Arithmetic sequence is calculate from

 Sn = n/2[a + l], where a is the first term and l is the last term

* Now lets solve the problem

- We will use method (1)

- From the table the terms of the sequence are:

 10 , 22 , 34 , 46 , 58 , 82 , 94 , ............., where 10 is the first term

∵ an = a1 + (n - 1) d ⇒ explicit formula

∵ a1 = 10 and a2 = 22

∵ d = a2 - a1

∴ d = 22 - 10 = 12

- The 100th term means the term of n = 100

∴ a100 = 10 + (100 - 1) 12

∴ a100 = 10 + 99 × 12 = 10 + 1188 = 1198

∴ The 100th term is 1198

- Lets find the sum of the first 100 terms of the sequence

∵ Sn = n/2[a1 + an]

∵ n = 100 , a = 10 , a100 = 1198

∴ S100 = 100/2[10 + 1198] = 50[1208] = 60400

* The sum of the first 100 terms is 60400

7 0
3 years ago
Solve for x. enter your answer in interval notation using grouping symbols. x^2+15x<-36
Oliga [24]
X²+15x+36<0

at first solve quadratic equation

D=b²-4ac= 225-4*1*36= 81

x=(-b+/-√D)/2a
x=(-15+/-√81)/2= (-15+/-9)/2
x1=(-15-9)/2=-12
x2=(-15+9)/2=-3

we can write x²+15x+36<0 as (x+12)(x+3)<0

(x+12)(x+3)<0 can be 2 cases, because for product to be negative one factor should be negative , and second factor should be positive
 1 case)      x+12<0, and x+3>0,                                            
                       x<-12, and x>-3
(-∞, -12) and(-3,∞) gives empty set

or second case)  x+12>0 and x+3<0
x>-12 and x<-3
(-12,∞) and (-∞,-3)  they are crossing , so (-12, -3)  is a solution of this inequality


3 0
4 years ago
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