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
Answer with Step-by-step explanation:
Let a mass weighing 16 pounds stretches a spring
feet.
Mass=
Mass=

Mass,m=
Slug
By hook's law




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

Equation of motion :

Using this equation



Auxillary equation





Complementary function

To find the particular solution using undetermined coefficient method



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


Comparing on both sides


Adding both equation then, we get


Substitute the value of B in any equation



Particular solution, 
Now, the general solution

From initial condition
x(0)=2 ft
x'(0)=0
Substitute the values t=0 and x(0)=2




Substitute x'(0)=0




Substitute the values then we get

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
Answer:
<em>- 7n ≤ 77</em>
Step-by-step explanation:
<em>- 7n ≤ 77</em>
n ≥ - 11
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
and standard deviation
, the zscore of a measure X is given by:

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:

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 




Upper limit
X when Z has a pvalue of 0.80. So X when 



