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3 years ago

Given Q is between J and K, JQ is five times the length of JK and JK=42 find the length of JQ

Mathematics

1 answer:

Alex73 [517]3 years ago

8 0

Answer:

210

Step-by-step explanation:

JQ is 5 times the length of JK

● JQ = 5JK

JK = 42

● JQ = 5 × 42

● JQ = 210

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Equation of the line with the given slope and point slope= -5/8 given point= (3,2)

avanturin [10]

Answer:

x =3

y = 2

y= 15/8 (3) + 2

Step-by-step explanation:

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3 years ago

Tyler's mom purchased a savings bond for Tyler. The value of the savings bond increases by 4% each year. One year after it was p

Gemiola [76]

Answer:

The value of the bond when Tyler's mom purchased it was $150

Step-by-step explanation:

we know that

In this problem we have a exponential function of the form

$f(x)=a(b)^{x}$

where

a is the initial value (y-intercept)

b is the base

r is the rate

b=(1+r)

In this problem

r=4%=4/100=0.04

b=1+0.04=1.04

substitute

$f(x)=a(1.04)^{x}$

where

x is the number of years since the savings bond was purchased

f(x) is the value of the savings bond

For x=1

f(x)=$156

substitute

$156=a(1.04)^{1}$

Solve for a

$156=a(1.04)$

$a=\$150$

therefore

The value of the bond when Tyler's mom purchased it was $150

8 0

3 years ago

Function A is represented by the equation

Dafna1 [17]

Answer:

They have the same rate of change.

Step-by-step explanation:

Well the graph of Function B is y = 1 / 3x + 2

So they have the same rate of change, since they have the same slope.

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3 years ago

Alaina was asked whether the following equation is an identity:

jarptica [38.1K]

Answer: yerror she made is at the level of the first step if you want me to show it to you ask me in comment

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2 years ago

Read 2 more answers

i need a box with a square base and an open top that cna hold 500 in^3. what is the least amount of material i can use to build

exis [7]

Answer:

M(x) = 452,56 in

Step-by-step explanation:

The volume of the open box is 500 in³

V = Area of the base times height

V(b) = x² * h where x is the side of the square and h the heigh

Then 500 = x²*h

Total material to use is: material of the base + material of 4 sides

material of the base is x²

material of one side is x*h we have 4 sides then 4*x*h

Total material M(b)

M(b) = x² + 4*x*h

And as h = 500/ x²

M(x) = x² + 4* x* 500/x²

M(x) = x² + 2000/x

Tacking derivatives on both sides of the equation

M´(x) = 2*x - 200/x²

M´(x) = 0 2*x - 200/x² = 0

x³ - 100 = 0

x³ = 100

x = 4,64 in

And By substitution h = 500/x²

h = 500/(4,64)² h = 23,22 in

How do we know that x = 4,64 make V(x) minimum

we get the second derivative

M´´(x) = 2 + 200*2x/ x⁴ = 2 + 400/x³

M´´(x) is always positive

M´´(x) > 0 then M(x) has a minimum for x= 4,64 in

The least amount of material is:

M(x) = x² + 2000/x

M(x) = (4,64)² * 2000/4,64

M(x) = 21,53 + 431,03

M(x) = 452,56 in

8 0

3 years ago