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Irina18 [472]
3 years ago
13

Evaluate the following at select the best answer 8 (3 exponent seven) minus (6 + 4) 2​

Mathematics
1 answer:
photoshop1234 [79]3 years ago
3 0

Answer:

17476

Step-by-step explanation:

8(3^{7})-(6+4)2

8(2187)-(10)2

17496-20

17476

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To have $25,000 to spend on a new car in five years, how much money should Jill invest today at 8% compounded monthly?
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The answer is B
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7 0
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VERY URGENT PLEASE HELP!!!!
Triss [41]

Answer:

61 + 4x = y

Step-by-step explanation:

61 : starting tree height

4x : 4 times however many months pass

y : total height after x months.

4 0
2 years ago
Find the slope between the two points given. Then, use the slope and Point One to write the equation of the line in Point-Slope
ratelena [41]

The equation in point slope form is given as y+3 = -5/8(x-4)

<h3>Equation of a line</h3>

The formula for calculating the equation of a line in point-slope form is expressed as:

y-y1= m(x-x1)

Given the coordinate point

Slope = 2-(-3)/-2-4
Slope = 5/-8

Determine the equation

y-(-3) =-5/8(x-4)

y+3 = -5/8(x-4)

Hence the equation in point slope form is given as y+3 = -5/8(x-4)

Learn more on equation of a line here: brainly.com/question/18831322

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5 0
1 year ago
(Algebra 1 )please help me I'll give brainliest
riadik2000 [5.3K]

7-3/-3-3

4/-6

this can be simplified to -2/3

(3) = -2/3(3) + b

Solve this

3 = -2 + b

5 = b

The answer is

y = -2/3x + 5

7 0
2 years ago
Solve y'' + 10y' + 25y = 0, y(0) = -2, y'(0) = 11 y(t) = Preview
svetlana [45]

Answer:  The required solution is

y=(-2+t)e^{-5t}.

Step-by-step explanation:   We are given to solve the following differential equation :

y^{\prime\prime}+10y^\prime+25y=0,~~~~~~~y(0)=-2,~~y^\prime(0)=11~~~~~~~~~~~~~~~~~~~~~~~~(i)

Let us consider that

y=e^{mt} be an auxiliary solution of equation (i).

Then, we have

y^prime=me^{mt},~~~~~y^{\prime\prime}=m^2e^{mt}.

Substituting these values in equation (i), we get

m^2e^{mt}+10me^{mt}+25e^{mt}=0\\\\\Rightarrow (m^2+10y+25)e^{mt}=0\\\\\Rightarrow m^2+10m+25=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow m^2+2\times m\times5+5^2=0\\\\\Rightarrow (m+5)^2=0\\\\\Rightarrow m=-5,-5.

So, the general solution of the given equation is

y(t)=(A+Bt)e^{-5t}.

Differentiating with respect to t, we get

y^\prime(t)=-5e^{-5t}(A+Bt)+Be^{-5t}.

According to the given conditions, we have

y(0)=-2\\\\\Rightarrow A=-2

and

y^\prime(0)=11\\\\\Rightarrow -5(A+B\times0)+B=11\\\\\Rightarrow -5A+B=11\\\\\Rightarrow (-5)\times(-2)+B=11\\\\\Rightarrow 10+B=11\\\\\Rightarrow B=11-10\\\\\Rightarrow B=1.

Thus, the required solution is

y(t)=(-2+1\times t)e^{-5t}\\\\\Rightarrow y(t)=(-2+t)e^{-5t}.

6 0
3 years ago
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