Find the values of k for which y = kx + 1 is a tangent to the curve y = 2x^2 + x +3

a positive number is 7 times another number if 3 is added to both the numbers then one of the new number becomes 5 by 2 times th

Neko [114]

Answer:

7 and 1

Step-by-step explanation:

Let the numbers be a and b.

A positive number is 7 times another number:

a = 7b

If 3 is added to both the numbers then one of the new number becomes 5 by 2 times the other new number:

a+3 = 5/2 × (b +3)

To solve this we substitute a with 7b in the second equation:

7b + 3 = 5/2 × (b +3) ⇒ multiplying both sides by 2
14b + 6 = 5b + 15 ⇒ collecting like terms
14b - 5b = 15 - 6
9b = 9
b = 1 ⇒ solved for b

Then, finding a:

a= 7b
a=7*1
a= 7 ⇒ solved for a

So the numbers are 7 and 1

3 0

3 years ago

Between what pair of numbers is the product of 289 and 7

LiRa [457]

The product of 289 and 7 is somewhere between 289 and 2,100 .

6 0

3 years ago

A past study claims that adults in America spend an average of 17 hours a week on leisure activities. A researcher wanted to tes

8090 [49]

Answer:

The claim is false.

Step-by-step explanation:

Given the data :

13 24 21 37 15 25 18 22 40 32

The sample mean and standard deviation can ben calculated for the given sample.

Using calculator :

Sample mean, xbar = 24.7

Sample standard deviation, s = 9.04

Sample size, n = 10

The hypothesis :

H0 : μ = 17

H1 : μ ≠ 17

The test statistic :

(xbar - μ) ÷ (s/√(n))

(24.7 - 17) ÷ (9.04/√(10))

7.7 / 2.8586990

Test statistic = 2.694

We can obtain the Pvalue, at α = 0.05 ; df = n-1 = 9

Pvalue = 0.0246

Since Pvalue < α ; we reject the null ; Hence, there is significant evidence to conclude that an adult American does not spend average of 17 hours in leisure

5 0

3 years ago

At noon, ship A is 130 km west of ship B. Ship A is sailing east at 25 km/h and ship B is sailing north at 20 km/h. How fast is

Hitman42 [59]

Answer:

answer = 12.87 km/h

Step-by-step explanation:

Given

Ship A is sailing east at 25 km/h = $\frac{dx}{dt}$

ship B is sailing north at 20 km/h = $\frac{dy}{dt}$

here x and y are the sailing at t = 4 : 00 pm for ship A and B respectively

so we get x = 4 ×25 =100 km/h

y = 4× 20 = 80 km/h

let z is the distance between the ships, we need to find $\frac{dz}{dt}$ at t = 4 hr

At noon, ship A is 130 km west of ship B (12:00 pm)

so equation will be

$z^2 = (130-x)^2 + y^2......................(i)\\put x = 100 and y = 80 \\\\we | | get \\$

$z^2 = 30^2 + 80^2\\z =\sqrt{7300} km/h$

derivative first equation w . r. to t we get

$2z\frac{dz}{dt} =$ $-2(130-x)\frac{dx}{dt}$ $+2y\frac{dy}{dt}$

$\frac{dz}{dt} =\frac{1}{z}[(x -130)\frac{dx}{dt} +y\frac{dy}{dt}]$

$\frac{dz}{dt} = \frac{( -20\times25 + 80\times20)}{\sqrt{7300} }$

$= \frac{1100}{85.44}\\ = 12.87km/h$

8 0

3 years ago

Help me with this please! :D

Nady [450]

Answer:

#1=2

#2= 14

#3= 26

#4= 32

Step-by-step explanation:

y=6x-16

Lets use the first one as an example: since x is our variable we can plug in 3 into the equation to make y=(6x3)-16. 6x3=18, so y=18-16 which is 2. You can plug in any other number to find your answer as long as you follow the pattern in the equation.

8 0

3 years ago

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