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Gnom [1K]
3 years ago
5

What is 12*12+100-50/200%*450

Mathematics
1 answer:
Levart [38]3 years ago
7 0

Answer:

-11006

Step-by-step explanation:

the answer is -11006

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-3x-9=-15 solve for x​
qaws [65]

Answer:x=2

Step-by-step explanation:

-3x-9=-15

-3x=-6

x=2

7 0
3 years ago
Read 2 more answers
So im really confused by this and could use some help.
Otrada [13]

Answer:

Nancy fits the equation.

Step-by-step explanation:

I think they're asking which gorilla fits the equation.

2b - 15 = 605

<em>Add 15 to both sides</em>

2b=620

<em>Divide both sides by 2</em>

b=310

The only gorilla who weighs 310 pounds is Nancy.

5 0
4 years ago
For any positive numbers a, b, and d, with b 1, logb(a d) = _____.
TEA [102]
For example:
log _{2} 20 = log _{2} ( 4 * 5 ) = \\ = log _{2} 4 + log _{2} 5
log_{b}( a d ) = log _{b}a +log _{b}d
Answer:
C ) log(b) a + log(b) d
8 0
3 years ago
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PLEASE SOME MATH HELP PLEASE................
Gnoma [55]
The measurement for The measurement for arc AC is 99.8
6 0
3 years ago
Solve y" + y = tet, y(0) = 0, y'(0) = 0 using Laplace transforms.
irina1246 [14]

Answer:

The solution of the diferential equation is:

y(t)=\frac{1}{2}cos(t)- \frac{1}{2}e^{t}+\frac{t}{2} e^{t}

Step-by-step explanation:

Given y" + y = te^{t}; y(0) = 0 ; y'(0) = 0

We need to use the Laplace transform to solve it.

ℒ[y" + y]=ℒ[te^{t}]

ℒ[y"]+ℒ[y]=ℒ[te^{t}]

By using the Table of Laplace Transform we get:

ℒ[y"]=s²·ℒ[y]+s·y(0)-y'(0)=s²·Y(s)

ℒ[y]=Y(s)

ℒ[te^{t}]=\frac{1}{(s-1)^{2}}

So, the transformation is equal to:

s²·Y(s)+Y(s)=\frac{1}{(s-1)^{2}}

(s²+1)·Y(s)=\frac{1}{(s-1)^{2}}

Y(s)=\frac{1}{(s^{2}+1)(s-1)^{2}}

To be able to separate in terms, we use the partial fraction method:

\frac{1}{(s^{2}+1)(s-1)^{2}}=\frac{As+B}{s^{2}+1} +\frac{C}{s-1}+\frac{D}{(s-1)^2}

1=(As+B)(s-1)² + C(s-1)(s²+1)+ D(s²+1)

The equation is reduced to:

1=s³(A+C)+s²(B-2A-C+D)+s(A-2B+C)+(B+D-C)

With the previous equation we can make an equation system of 4 variables.

The system is given by:

A+C=0

B-2A-C+D=0

A-2B+C=0

B+D-C=1

The solution of the system is:

A=1/2 ; B=0 ; C=-1/2 ; D=1/2

Therefore, Y(s) is equal to:

Y(s)=\frac{s}{2(s^{2} +1)} -\frac{1}{2(s-1)} +\frac{1}{2(s-1)^{2}}

By using the inverse of the Laplace transform:

ℒ⁻¹[Y(s)]=ℒ⁻¹[\frac{s}{2(s^{2} +1)}]-ℒ⁻¹[\frac{1}{2(s-1)}]+ℒ⁻¹[\frac{1}{2(s-1)^{2}}]

y(t)=\frac{1}{2}cos(t)- \frac{1}{2}e^{t}+\frac{t}{2} e^{t}

8 0
3 years ago
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