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2 years ago

Can you pls help me thank you have a good day

Mathematics

2 answers:

KengaRu [80]2 years ago

8 0

Answer:

y = 2x + 3

Step-by-step explanation:

3 is the constant

2 is the rate at which recieves

So, y = 2x + 3

Westkost [7]2 years ago

8 0

Answer:

Step-by-step explanation:

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Step-by-step explanation:

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2 years ago

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1 year ago

MARKING BRAINLIEST PLZ HURRY<br><br> 1 1/4 + ? = 0

Rzqust [24]

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2 years ago

PLS, HELP WILL GIVE BRAINIEST THX AND 30 POINTS 4 BEST ANSWER Jenna just opened a checking account at the bank. Within the first

nadya68 [22]

A) Deposits: +34.98; +51.02; +51.22
Withdrawals: -5.23; -5.22; -7.89
B) -7.89, -5.23, -5.22, +34.98, +51.02, +51.22
The smallest amount will be the largest withdrawal, as it takes away from the account; the largest amount will be the largest deposit, as it adds to the account.
C) She would have $216.88.

98 + 34.98 = 132.98 + 51.02 = 184.00 + 51.22 = 235.22 - 7.89 = 227.33 - 5.23 = 222.10 - 5.22 = 216.88

6 0

2 years ago

F(x, y, z) = yzexzi + exzj + xyexzk,

Ghella [55]

$\nabla f(x,y,z)=\mathbf F(x,y,z)\iff\dfrac{\partial f}{\partial x}\,\mathbf i+\dfrac{\partial f}{\partial y}\,\mathbf j+\dfrac{\partial f}{\partial z}\,\mathbf k=yze^{xz}\,\mathbf i+e^{xz}\,\mathbf j+xye^{xz}\,\mathbf k$

$\dfrac{\partial f}{\partial x}=yze^{xz}$
$\implies f(x,y,z)=\dfrac{yz}ze^{xz}+g(y,z)=ye^{xz}+g(y,z)$

$\dfrac{\partial f}{\partial y}=e^{xz}=e^{xz}+\dfrac{\partial g}{\partial y}$
$\implies\dfrac{\partial g}{\partial y}=0\implies g(y,z)=h(z)$

$\dfrac{\partial f}{\partial z}=xye^{xz}=xye^{xz}+\dfrac{\mathrm dh}{\mathrm dz}$
$\implies\dfrac{\mathrm dh}{\mathrm dz}=0\implies h(z)=C$

$f(x,y,z)=ye^{xz}+C$

$\mathbf r(t)=(t^2+5)\,\mathbf i+(t^2-1)\,\mathbf j+(t^2-5)\,\mathbf k$
$\implies\mathbf r(0)=5\,\mathbf i-\mathbf j-5\,\mathbf k$
$\implies\mathbf r(5)=30\,\mathbf i+24\,\mathbf j+20\,\mathbf k$

By the gradient theorem, we have for any path $\mathcal C$ originating at the point (5, -1, -5) and terminating at the point (30, 24, 20),

$\displaystyle\int_{\mathcal C}\mathbf F\cdot\mathrm d\mathbf r=f(\mathbf r(5))-f(\mathbf r(0))=f(30,24,20)-f(5,-1,-5)$
$\implies\displaystyle\int_{\mathcal C}\mathbf F\cdot\mathrm d\mathbf r=24e^{600}+e^{-25}$

4 0

2 years ago