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boyakko [2]
3 years ago
15

Which equation shows P=2l + 2w when solved for w?

Mathematics
2 answers:
larisa86 [58]3 years ago
7 0

Answer:

Is d. W=P-2L/2

Jlenok [28]3 years ago
3 0

Answer:

the answer is A

Step-by-step explanation:

when solving for a literal equation you wanna get the letter by itself

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Please answer the question and please answer correctly... I really need help
Mekhanik [1.2K]

Answer:

The treatment decays by half each week.

Step-by-step explanation:

The decay function is given by:

y=a(1-r)^{t}

Here,

<em>y</em> = final value

<em>a</em> = initial value

<em>r</em> = decay rate

<em>t </em>= time

The decay function for one medications to treat dogs for fleas is:

F(x)=4\cdot (\frac{1}{2})^{w}

Here the decay rate is:

1-r=\frac{1}{2}\\\\r=1-\frac{1}{2}\\\\r=\frac{1}{2}

The treatment decays by half each week.

3 0
3 years ago
Name the set of 4 consecutive integers starting with -4
AVprozaik [17]
-4, -3, -2, -1

Consecutive integers are integers that are exactly one unit away from each other. -4 is one unit away from -3, and likewise with -3 and -2 and -2 and -1. 
7 0
3 years ago
Find constants a and b such that the function y = a sin(x) + b cos(x) satisfies the differential equation y'' + y' − 5y = sin(x)
vichka [17]

Answers:

a = -6/37

b = -1/37

============================================================

Explanation:

Let's start things off by computing the derivatives we'll need

y = a\sin(x) + b\cos(x)\\\\y' = a\cos(x) - b\sin(x)\\\\y'' = -a\sin(x) - b\cos(x)\\\\

Apply substitution to get

y'' + y' - 5y = \sin(x)\\\\\left(-a\sin(x) - b\cos(x)\right) + \left(a\cos(x) - b\sin(x)\right) - 5\left(a\sin(x) + b\cos(x)\right) = \sin(x)\\\\-a\sin(x) - b\cos(x) + a\cos(x) - b\sin(x) - 5a\sin(x) - 5b\cos(x) = \sin(x)\\\\\left(-a\sin(x) - b\sin(x) - 5a\sin(x)\right)  + \left(- b\cos(x) + a\cos(x) - 5b\cos(x)\right) = \sin(x)\\\\\left(-a - b - 5a\right)\sin(x)  + \left(- b + a - 5b\right)\cos(x) = \sin(x)\\\\\left(-6a - b\right)\sin(x)  + \left(a - 6b\right)\cos(x) = \sin(x)\\\\

I've factored things in such a way that we have something in the form Msin(x) + Ncos(x), where M and N are coefficients based on the constants a,b.

The right hand side is simply sin(x). So we want that cos(x) term to go away. To do so, we need the coefficient (a-6b) in front of that cosine to be zero

a-6b = 0

a = 6b

At the same time, we want the (-6a-b)sin(x) term to have its coefficient be 1. That way we simplify the left hand side to sin(x)

-6a  -b = 1

-6(6b) - b = 1 .... plug in a = 6b

-36b - b = 1

-37b = 1

b = -1/37

Use this to find 'a'

a = 6b

a = 6(-1/37)

a = -6/37

8 0
3 years ago
Point-slope of line's equation (-3,-7),(-8-11)
Nataly_w [17]
Steps:
1. Find the slope by using the slope formula
Y2-Y1
---------
X2-X1

-11 + 7
----------
-8 + 3
2.
-11 + 7= -4
-8 + 3= -5
3. Your slope is 4/5 (because you have to turn two negatives into a positive)
4. Now you do the point-slope formula
Y-Y1= m(x-x1)
Y +7=4/5 (x +3)
3 0
3 years ago
Using mathematical induction prove whether or not the following statement is true for all positive integers n, or show why it is
aniked [119]

Base case: For n=1, the left side is 2 and the right is 2\cdot1^2=2, so the base case holds.

Induction hypothesis: Assume the statement is true for n=k, that is

2+6+10+\cdots+4k-2=2k^2

We want to show that this implies truth for n=k+1, that

2+6+10+\cdots+4k-2+4(k+1)-2=2(k+1)^2

The first k terms on the left reduce according to the assumption above, and we can simplify the k+1-th term a bit:

\underbrace{2+6+10+\cdots+4k-2}_{2k^2}+4k+2

2k^2+4k+2=2(k^2+2k+1)=2(k+1)^2

so the statement is true for all n\in\mathbb N.

5 0
4 years ago
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