suppose Monica walks 1 kilometer every 15 minutes how many meters further can she walk in 1 hour at this new rate

Answers

Answer 1
Answer: if Monica walks
1km...........................15min
?km............................60min
60/15=4
Monica walks 4km in 60 min(1H)
1km=1,000m
4km=4,000m in 1 hour

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A ball is dropped from a height if 30 feet the ball bounces after each bounce the height attained by the ball is 60% of the previous height write an nth term formula to model the situation what is the maximum height attained by the ball after five bouncesA.0.12ft
B.6.40ft
C.2.33ft
D.1.40ft

Answers

This information will be modeled using the formula
thus we shall have:
Sn=ar^n
where:
a=first term
r=common ratio
from the information:
a=30 ft
r=60/100=3/5=0.6
therefore the formula will be
Sn=30(0.6)^n
where n is the number of terms:
thus when n=5 th sum will be:
S5=30(0.6)^5
S5=30(0.6)^5
S5=2.33 ft
Answer: 2.33 ft

Answer:

2.33

Step-by-step explanation:

Convert 0.49494949 into a fraction. Show all work.

Answers

Answer:

0.4949

=

49      this is a fraction

99

Step-by-step explanation:

Let

XXX

x

=

0.49

¯¯¯¯

49

then

XXX

100

x

=

49.49

¯¯¯¯

49

and

XXX

99

x

=

100

x

x

=

49

XXX

x

=

49

99

How do you do these two questions?

Answers

Answer:

(a) ⅛ tan⁻¹(¼)

(b) sec x − ln│csc x + cot x│+ C

Step-by-step explanation:

(a) ∫₀¹ x / (16 + x⁴) dx

∫₀¹ (x/16) / (1 + (x⁴/16)) dx

⅛ ∫₀¹ (x/2) / (1 + (x²/4)²) dx

If tan u = x²/4, then sec²u du = x/2 dx

⅛ ∫ sec²u / (1 + tan²u) du

⅛ ∫ du

⅛ u + C

⅛ tan⁻¹(x²/4) + C

Evaluate from x=0 to x=1.

⅛ tan⁻¹(1²/4) − ⅛ tan⁻¹(0²/4)

⅛ tan⁻¹(¼)

(b) ∫ (sec³x / tan x) dx

Multiply by cos x / cos x.

∫ (sec²x / sin x) dx

Pythagorean identity.

∫ ((tan²x + 1) / sin x) dx

Divide.

∫ (tan x sec x + csc x) dx

Split the integral

∫ tan x sec x dx + ∫ csc x dx

Multiply second integral by (csc x + cot x) / (csc x + cot x).

∫ tan x sec x dx + ∫ csc x (csc x + cot x) / (csc x + cot x) dx

Integrate.

sec x − ln│csc x + cot x│+ C

Answer:

(a) Solution : 1/8 cot⁻¹(4) or 1/8 tan⁻¹(¼) (either works)

(b) Solution : tan(x)/sin(x) + In | tan(x/2) | + C

Step-by-step explanation:

(a) We have the integral (x/16 + x⁴)dx on the interval [0 to 1].

For the integrand x/6 + x⁴, simply pose u = x², and du = 2xdx, and substitute:

1/2 ∫ (1/u² + 16)du

'Now pose u as 4v, and substitute though integral substitution. First remember that we have to factor 16 from the denominator, to get 1/2 ∫ 1/(16(u²/16 + 1))' :

∫ 1/4(v² + 1)dv

'Use the common integral ∫ (1/v² + 1)dv = arctan(v), and substitute back v = u/4 to get our solution' :

1/4arctan(u/4) + C

=> Solution : 1/8 cot⁻¹(4) or 1/8 tan⁻¹(¼)

(b) We have the integral ∫ sec³(x)/tan(x)dx, which we are asked to evaluate. Let's start by substitution tan(x) as sin(x)/cos(x), if you remember this property. And sec(x) = 1/cos(x) :

∫ (1/cos(x))³/(sin(x)/cos(x))dx

If we cancel out certain parts we receive the simplified expression:

∫ 1/cos²(x)sin(x)dx

Remember that sec(x) = 1/cos(x):

∫ sec²(x)/sin(x)dx

Now let's start out integration. It would be as follows:

\mathrm{Let:u=(1)/(\sin \left(x\right)),\:v'=\sec ^2\left(x\right)}\n=> (\tan \left(x\right))/(\sin \left(x\right))-\int \:-\cot \left(x\right)\csc \left(x\right)\tan \left(x\right)dx\n\n\int \:-\cot \left(x\right)\csc \left(x\right)\tan \left(x\right)dx=-\ln \left|\tan \left((x)/(2)\right)\right|\n=> (\tan \left(x\right))/(\sin \left(x\right))-\left(-\ln \left|\tan \left((x)/(2)\right)\right|\right)\n

=> (\tan \left(x\right))/(\sin \left(x\right))+\ln \left|\tan \left((x)/(2)\right)\right|\n\n=> (\tan \left(x\right))/(\sin \left(x\right))+\ln \left|\tan \left((x)/(2)\right)\right|+C

Solution: tan(x)/sin(x) + In | tan(x/2) | + C

When does a quadratic equation have no solutions

Answers

Answer:

A quadratic equation has solutions when the graph crosses the x-axis. There are two ways the graph can have no solution, when the "a" value is greater than 0 and is translated vertically above the x-axis, or if the opposite occurs, when the "a" value is negative and is translated vertically below the x-axis.

Which equation represents a proportional relationship that has a constant of proportionality equal to One-fifth?

Answers

Answer:

y=1/5x  or y=.2x

Step-by-step explanation:

the constant=k the equation is always  y=kx  and if the constant of proportionality is 1/5 you would do  y=1/5x  or y=.2x

Answer:

C. y = one-fifth x

Step-by-step explanation:

I took the quiz

On Wednesday, a local hamburger shop sold a combined total of 360 hamburgers and cheeseburgers. The number of cheeseburgers sold was two times the number of hamburgers sold. How many hamburgers were sold on Wednesday?

Answers

Answer:

120

Step-by-step explanation:

360 combined, 2/3 is cheese, so 1/3 is hamburger, 1/3 of 360 is 120.