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2 years ago

-17 = 6p - 5 What is p?

Mathematics

2 answers:

max2010maxim [7]2 years ago

5 0

The answer is p=-2 because you add 5 to -17 and you get-12 then you do 6/-12 and you get -2

Galina-37 [17]2 years ago

3 0

Answer:

p =-2

Step-by-step explanation:

Add 5 to both sides: -17(+5)=6p-5(+5)

Divide both sides by 6: -12/6=6p/6

Simplify: p=-2.

Hope this helps! :)

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Identify the DELETION mutation from the following DNA sequence:ATG CCA AAT

Umnica [9.8K]

Answer

Hello, There!

:Identify the DELETION mutation from the following DNA sequence

ATG CCA AAT.

Hope this helps!

4 0

2 years ago

How did the temperature change if: At first it increased by 25% and then decreased by 40%

Roman55 [17]

Answer:

24.6 % decrease

Step-by-step explanation:

Let x be the original temperature

Increase by 25%

x + .25x

1.25x

Decrease by 40%

The amount decreased is

(1.25x) *.40

.496x

Subtract that from 1.25x

1.25x - .496x

.754x

We need to find the percent decrease

1 - .754 = .246

24.6 % decrease

7 0

2 years ago

Read 2 more answers

For each of the following vector fields F , decide whether it is conservative or not by computing curl F . Type in a potential f

Phantasy [73]

The key idea is that, if a vector field is conservative, then it has curl 0. Equivalently, if the curl is not 0, then the field is not conservative. But if we find that the curl is 0, that on its own doesn't mean the field is conservative.

$\mathrm{curl}\vec F=\dfrac{\partial(5x+10y)}{\partial x}-\dfrac{\partial(-6x+5y)}{\partial y}=5-5=0$

We want to find $f$ such that $\nabla f=\vec F$ . This means

$\dfrac{\partial f}{\partial x}=-6x+5y\implies f(x,y)=-3x^2+5xy+g(y)$

$\dfrac{\partial f}{\partial y}=5x+10y=5x+\dfrac{\mathrm dg}{\mathrm dy}\implies\dfrac{\mathrm dg}{\mathrm dy}=10y\implies g(y)=5y^2+C$

$\implies\boxed{f(x,y)=-3x^2+5xy+5y^2+C}$

so $\vec F$ is conservative.

$\mathrm{curl}\vec F=\left(\dfrac{\partial(-2y)}{\partial z}-\dfrac{\partial(1)}{\partial y}\right)\vec\imath+\left(\dfrac{\partial(-3x)}{\partial z}-\dfrac{\partial(1)}{\partial z}\right)\vec\jmath+\left(\dfrac{\partial(-2y)}{\partial x}-\dfrac{\partial(-3x)}{\partial y}\right)\vec k=\vec0$

Then

$\dfrac{\partial f}{\partial x}=-3x\implies f(x,y,z)=-\dfrac32x^2+g(y,z)$

$\dfrac{\partial f}{\partial y}=-2y=\dfrac{\partial g}{\partial y}\implies g(y,z)=-y^2+h(y)$

$\dfrac{\partial f}{\partial z}=1=\dfrac{\mathrm dh}{\mathrm dz}\implies h(z)=z+C$

$\implies\boxed{f(x,y,z)=-\dfrac32x^2-y^2+z+C}$

so $\vec F$ is conservative.

$\mathrm{curl}\vec F=\dfrac{\partial(10y-3x\cos y)}{\partial x}-\dfrac{\partial(-\sin y)}{\partial y}=-3\cos y+\cos y=-2\cos y\neq0$

so $\vec F$ is not conservative.

$\mathrm{curl}\vec F=\left(\dfrac{\partial(5y^2)}{\partial z}-\dfrac{\partial(5z^2)}{\partial y}\right)\vec\imath+\left(\dfrac{\partial(-3x^2)}{\partial z}-\dfrac{\partial(5z^2)}{\partial x}\right)\vec\jmath+\left(\dfrac{\partial(5y^2)}{\partial x}-\dfrac{\partial(-3x^2)}{\partial y}\right)\vec k=\vec0$