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Rama09 [41]
4 years ago
10

Luisa is 43 feet underground touring a cavern. She climbs a ladder up 14feet. What was her location

Mathematics
1 answer:
Rzqust [24]4 years ago
7 0

Answer:

29 ft

Step-by-step explanation:

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What is the smallest perimeter possible for a rectangle whose area is 16in^2 and what are its dimensions? ...?
const2013 [10]
Area of the rectangle is given as = 16 square inches
We know
Length * width = Area
Then
Width = Area/Length
Now
Perimeter = 2 (Length + Width)
Perimeter = 2[ {Length + (Area/Length)}]
Perimeter = 2[Length + (16/Length)]
When d(Perimeter)/d(length) = 0,
Then
2[Length + (16/Length)] = 0
Length + (16/Length) = 0
Length^2 + 16 = 0
(Length)^2 = -(4)^2
Then
Length = 4 inches
Now
Width = Area/Length
          = 16/4
          = 4 inches.
4 0
3 years ago
Question 1*
worty [1.4K]

Given expressions;

     f(x) = x² + 6x + 7

     g(x) = 8x² - 2

     h(x) = 3x + 4

     k(x) = x - 3

Find  f(2) - g(2). k(2)

f(2) = 2² + 6(2) + 7 = 23

g(2) = 8(2)² - 2 = 30

k(2) = 2 - 3 = -1

Now,  f(2) - g(2). k(2) = 23  - (30)(-1)

                                  = 53

The solution to the problem is 53

6 0
4 years ago
Name the property shown by the statement. (5 + m) + n = 5+ (m + n)
SVEN [57.7K]

Answer:

Associative property of addition

4 0
3 years ago
What is the x-intercept in this linear equation?<br> 2y - 4x = -12
34kurt
X=3-2y   is the answer
3 0
3 years ago
Read 2 more answers
Evaluate the integral. (remember to use absolute values where appropriate. Use c for the constant of integration.) 5 cot5(θ) sin
ozzi

I=5\int \frac{cos^{4}\theta }{sin\theta }\times cos\theta d\theta \\\\I=5\int \left ( 1-sin^{2}\theta  \right )^{2}\times \frac{cos\theta }{sin\theta }d\theta \\put\ \sin\theta =t\\\\dt=cos\theta d\theta \\\\I=5\int\frac{t^{4}+1-2t^{2}}{t}dt\ \ \ \ \ \ \ \ \ \ \because (a-b)^2=a^2+b^2-2ab\\\\I=5\left ( \int t^{3}dt + \int \frac{1}{t} -2\int t \right )dt

by using the integration formula

we get,

\\I=5\left ( \frac{t^{4}}{4} +logt -t^{2}\right )\\\\I=\frac{5}{4}t^{4}+5\log t-5t^{2}+c

now put the value of t=\sin\theta in the above equation

we get,

\int 5\cot^5\theta \sin^4\theta d\theta=\frac{5}{4}sin^{4}\theta+5\log \sin\theta - 5sin^{2} \theta+c

hence proved

7 0
3 years ago
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