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Morgarella [4.7K]
3 years ago
7

FUN FACT

Mathematics
2 answers:
frosja888 [35]3 years ago
7 0

Answer:

word

Step-by-step explanation:

9966 [12]3 years ago
6 0

Answer:

trueeee

Step-by-step explanation:

You might be interested in
A rectangular photograph is nine centimeters wide and fourteen centimeters long. The photograph is enlarged by increasing the le
koban [17]

Answer:

Width=13cm

Length=18cm

Step-by-step explanation:

Width of the Initial Photograph=9cm

Length of the Initial Photograph=14cm

If width and length are increased by an equal amount, say x

New Width=9+x

New Length=14+x

Area of the new photo is 108 square centimeters greater than the area of the original photo.

Area of Original Photo=14*9=126cm²

Aeea of the New Photo=126+108=234cm²

Therefore:

(9+x)(14+x)=234

126+9x+14x+x²-234=0

x²+23x-108=0

x²+27x-4x-108=0

x(x+27)-4(x+27)=0

(x+27)(x-4)=0

x+27=0 or x-4=0

x=-27 or x=4

Since x≠-27

x=4cm

The dimensions of the new photo are:

Width=9+x=9+4=13cm

Length=14+x=14+4=18cm

4 0
3 years ago
A mass weighing 16 pounds stretches a spring (8/3) feet. The mass is initially released from rest from a point 2 feet below the
mezya [45]

Answer with Step-by-step explanation:

Let a mass weighing 16 pounds stretches a spring \frac{8}{3} feet.

Mass=m=\frac{W}{g}

Mass=m=\frac{16}{32}

g=32 ft/s^2

Mass,m=\frac{1}{2} Slug

By hook's law

w=kx

16=\frac{8}{3} k

k=\frac{16\times 3}{8}=6 lb/ft

f(t)=10cos(3t)

A damping force is numerically equal to 1/2 the instantaneous velocity

\beta=\frac{1}{2}

Equation of motion :

m\frac{d^2x}{dt^2}=-kx-\beta \frac{dx}{dt}+f(t)

Using this equation

\frac{1}{2}\frac{d^2x}{dt^2}=-6x-\frac{1}{2}\frac{dx}{dt}+10cos(3t)

\frac{1}{2}\frac{d^2x}{dt^2}+\frac{1}{2}\frac{dx}{dt}+6x=10cos(3t)

\frac{d^2x}{dt^2}+\frac{dx}{dt}+12x=20cos(3t)

Auxillary equation

m^2+m+12=0

m=\frac{-1\pm\sqrt{1-4(1)(12)}}{2}

m=\frac{-1\pmi\sqrt{47}}{2}

m_1=\frac{-1+i\sqrt{47}}{2}

m_2=\frac{-1-i\sqrt{47}}{2}

Complementary function

e^{\frac{-t}{2}}(c_1cos\frac{\sqrt{47}}{2}+c_2sin\frac{\sqrt{47}}{2})

To find the particular solution using undetermined coefficient method

x_p(t)=Acos(3t)+Bsin(3t)

x'_p(t)=-3Asin(3t)+3Bcos(3t)

x''_p(t)=-9Acos(3t)-9sin(3t)

This solution satisfied the equation therefore, substitute the values in the differential equation

-9Acos(3t)-9Bsin(3t)-3Asin(3t)+3Bcos(3t)+12(Acos(3t)+Bsin(3t))=20cos(3t)

(3B+3A)cos(3t)+(3B-3A)sin(3t)=20cso(3t)

Comparing on both sides

3B+3A=20

3B-3A=0

Adding both equation then, we get

6B=20

B=\frac{20}{6}=\frac{10}{3}

Substitute the value of B in any equation

3A+10=20

3A=20-10=10

A=\frac{10}{3}

Particular solution, x_p(t)=\frac{10}{3}cos(3t)+\frac{10}{3}sin(3t)

Now, the general solution

x(t)=e^{-\frac{t}{2}}(c_1cos(\frac{\sqrt{47}t}{2})+c_2sin(\frac{\sqrt{47}t}{2})+\frac{10}{3}cos(3t)+\frac{10}{3}sin(3t)

From initial condition

x(0)=2 ft

x'(0)=0

Substitute the values t=0 and x(0)=2

2=c_1+\frac{10}{3}

2-\frac{10}{3}=c_1

c_1=\frac{-4}{3}

x'(t)=-\frac{1}{2}e^{-\frac{t}{2}}(c_1cos(\frac{\sqrt{47}t}{2})+c_2sin(\frac{\sqrt{47}t}{2})+e^{-\frac{t}{2}}(-c_1\frac{\sqrt{47}}{2}sin(\frac{\sqrt{47}t}{2})+\frac{\sqrt{47}}{2}c_2cos(\frac{\sqrt{47}t}{2})-10sin(3t)+10cos(3t)

Substitute x'(0)=0

0=-\frac{1}{2}\times c_1+10+\frac{\sqrt{47}}{2}c_2

\frac{\sqrt{47}}{2}c_2-\frac{1}{2}\times \frac{-4}{3}+10=0

\frac{\sqrt{47}}{2}c_2=-\frac{2}{3}-10=-\frac{32}{3}

c_2==-\frac{64}{3\sqrt{47}}

Substitute the values then we get

x(t)=e^{-\frac{t}{2}}(-\frac{4}{3}cos(\frac{\sqrt{47}t}{2})-\frac{64}{3\sqrt{47}}sin(\frac{\sqrt{47}t}{2})+\frac{10}{3}cos(3t)+\frac{10}{3}sin(3t)

8 0
3 years ago
What is the area of the cross section if the length is 12 cm and the width is 10 cm?
VARVARA [1.3K]

Answer:

120 cm²

Step-by-step explanation:

From the question;

  • Length is 12 cm
  • Width is 10 cm

We are required  to determine the area of the cross section;

  • Note that the cross section is the plane of a solid that remains constant through the solid.
  • In this case, the cross section is a rectangle whose dimensions are 12 cm by 10 cm.

But Area of a rectangle = Length × Width

Therefore;

Area of cross section = 12 cm × 10 cm

                                   = 120 cm²

Thus, area of the cross section is 120 cm²

6 0
3 years ago
Write and solve an inequality to fit each phrase; the product of -7 and a number is no more than 77
erastovalidia [21]

Answer:

<em>- 7n ≤ 77</em>

Step-by-step explanation:

<em>- 7n ≤ 77</em>

n ≥ - 11

7 0
3 years ago
For a medical study, a researcher wishes to select people in the middle 60% of the population based on blood pressure.
Finger [1]

Answer:

Lower limit: 113.28

Upper limit: 126.72

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 120, \sigma = 8

Middle 60%

So it goes from X when Z has a pvalue of 0.5 - 0.6/2 = 0.2 to X when Z has a pvalue of 0.5 + 0.6/2 = 0.8

Lower limit

X when Z has a pvalue of 0.20. So X when Z = -0.84

Z = \frac{X - \mu}{\sigma}

-0.84 = \frac{X - 120}{8}

X - 120 = -0.84*8

X = 113.28

Upper limit

X when Z has a pvalue of 0.80. So X when Z = 0.84

Z = \frac{X - \mu}{\sigma}

0.84 = \frac{X - 120}{8}

X - 120 = 0.84*8

X = 126.72

4 0
3 years ago
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