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Pani-rosa [81]
3 years ago
7

Please help me with this I’m stuck.

Mathematics
1 answer:
hichkok12 [17]3 years ago
8 0
I think you would do 3 divided by 2 which is 1.5, times 4 which would be 6, take that - 3 would be 3. Then you would do 5/3 which is 1.67 times 3 which is 5. you would then take 3 + 5 and that would get you 8. hopefully this helped!
You might be interested in
7th grade Ms. Takemoto wants to raise her math grade to a 95 to improve her chances of winning a math scholarship. Her math aver
malfutka [58]
No, 15% of 81 is 12.5 and 81+12.5 is only 93.5 and she needs at least a 95
8 0
3 years ago
A rectangular shoebox has a volume of 728 cubic inches.the base of the shoebox measures 8 inches by 6.5 inches.how long is the s
Flauer [41]

Answer:the answer is 14

Step-by-step explanation: first you want to multiply

8*6.5

=52

then, since you want to get x being what it takes to get to 728 you divide

728/52

14

hope this helps

Goog luck!

5 0
3 years ago
Somebody helpppppppppppoo
Marizza181 [45]

Answer:

40 degrees : )

4 0
2 years ago
Suppose that \nabla f(x,y,z) = 2xyze^{x^2}\mathbf{i} + ze^{x^2}\mathbf{j} + ye^{x^2}\mathbf{k}. if f(0,0,0) = 2, find f(1,1,1).
lesya [120]

The simplest path from (0, 0, 0) to (1, 1, 1) is a straight line, denoted C, which we can parameterize by the vector-valued function,

\mathbf r(t)=(1-t)(\mathbf i+\mathbf j+\mathbf k)

for 0\le t\le1, which has differential

\mathrm d\mathbf r=-(\mathbf i+\mathbf j+\mathbf k)\,\mathrm dt

Then with x(t)=y(t)=z(t)=1-t, we have

\displaystyle\int_{\mathcal C}\nabla f(x,y,z)\cdot\mathrm d\mathbf r=\int_{t=0}^{t=1}\nabla f(x(t),y(t),z(t))\cdot\mathrm d\mathbf r

=\displaystyle\int_{t=0}^{t=1}\left(2(1-t)^3e^{(1-t)^2}\,\mathbf i+(1-t)e^{(1-t)^2}\,\mathbf j+(1-t)e^{(1-t)^2}\,\mathbf k\right)\cdot-(\mathbf i+\mathbf j+\mathbf k)\,\mathrm dt

\displaystyle=-2\int_{t=0}^{t=1}e^{(1-t)^2}(1-t)(t^2-2t+2)\,\mathrm dt

Complete the square in the quadratic term of the integrand: t^2-2t+2=(t-1)^2+1=(1-t)^2+1, then in the integral we substitute u=1-t:

\displaystyle=-2\int_{t=0}^{t=1}e^{(1-t)^2}(1-t)((1-t)^2+1)\,\mathrm dt

\displaystyle=-2\int_{u=0}^{u=1}e^{u^2}u(u^2+1)\,\mathrm du

Make another substitution of v=u^2:

\displaystyle=-\int_{v=0}^{v=1}e^v(v+1)\,\mathrm dv

Integrate by parts, taking

r=v+1\implies\mathrm dr=\mathrm dv

\mathrm ds=e^v\,\mathrm dv\implies s=e^v

\displaystyle=-e^v(v+1)\bigg|_{v=0}^{v=1}+\int_{v=0}^{v=1}e^v\,\mathrm dv

\displaystyle=-(2e-1)+(e-1)=-e

So, we have by the fundamental theorem of calculus that

\displaystyle\int_C\nabla f(x,y,z)\cdot\mathrm d\mathbf r=f(1,1,1)-f(0,0,0)

\implies-e=f(1,1,1)-2

\implies f(1,1,1)=2-e

3 0
3 years ago
I have asked so many times and no one will answer. :(
andrew-mc [135]

Answer:

1) \frac{1}{2^7} ; 2) 11^7

Step-by-step explanation:

1) Using the Power of a Fraction Rule (\frac{a}{b} )^x = (\frac{a^x}{b^x} ),

(\frac{1}{2} )^7 = (\frac{1^7}{2^7} ), which can just be simplified to \frac{1}{2^7}.

2) Using the Negative Exponent Rule, a^{-x}  = \frac{1}{a^{x}},

(\frac{1}{11} )^{-7} = \frac{1}{\frac{1}{11^7}}, which can be simplified to 11^7.

4 0
2 years ago
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