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Naddika [18.5K]
3 years ago
14

A 100g packet of tea costs £4.16.

Mathematics
1 answer:
ANTONII [103]3 years ago
7 0

Answer:

A 100g packet of tea for £4.16

Step-by-step explanation:

If you multiply the second item, the 25g packet of tea for £1.05, by 4, it will come out to 100g for £4.20.

£4.20>£4.16; therefore, the 100g packet of tea is a better deal.

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The function f(x)=49x represents the number of jumping jacks f(x) can do in x minutes. how many jumping jacks can you do in 5 mi
lys-0071 [83]
245, if x represents the number of minutes then you use 5 where x would be in the function so 49(5) which is 245 jumping jacks in 5 minutes.
4 0
3 years ago
It is ask how many times did it say beetlejuice You could use a buddy
STALIN [3.7K]

Answer:

7 times

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Have a great day! Just know to ty even trough rough times<br><br><br><br>​
topjm [15]

Answer:

y....u.......p

Step-by-step explanation:

good day :D

6 0
2 years ago
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Consider f and c below. f(x, y, z) = yzexzi + exzj + xyexzk,
Jet001 [13]

Answer:

f(x,y,z)=ye^{xz}+C  

Step-by-step explanation:

We can write the given expression as :

\vec f(x,y,z)=yze^{xz}\,\vec\imath+e^{xz}\,\vec\jmath+xye^{xz}\,\vec k

As given,   f = ∇f.

∇f = \dfrac{\partial f}{\partial x}i  + \dfrac{\partial f}{\partial y}j  +\dfrac{\partial f}{\partial z}k 

We can write the partial derivative with respect to x, y and z.

\dfrac{\partial f}{\partial x}=yze^{xz}       ___(Equation 1)

\dfrac{\partial f}{\partial y}=e^{xz}               ______(Equation 2)

\dfrac{\partial f}{\partial z}=xye^{xz}             ______(Equation 3)

Take equation 2 and integrate with respect to y,

\dfrac{\partial f}{\partial y}=e^{xz}

f(x,y,z)=ye^{xz}+a(x,z)           ----------Equation 4

Derivate both sides w.r.t x , we get :

\frac{d}{dx}(yze^{xz})=yze^{xz}+\dfrac{\partial a}{\partial x}

or

\dfrac{\partial a}{\partial x}=0

integrate

a(x,z)=b(z)

put in equation 4 ,

we get :

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

take derivative wrt z

\frac{d}{dz} (ye^{xz}+b(z))\impliesxye^{xz}=xye^{xz}+\frac{db}{dz}

we can take here:

\frac{db}{dz} = 0

integrate:

\int\ {\frac{db}{dz} } \, =\int0

b(z) = C

The function can be written as :

from equation 4 :

f(x,y,z)=ye^{xz}+C  

Where C is a constant.

4 0
3 years ago
I need help, I need an explanation for the answer
hammer [34]

Answer:

3) The function's leading coefficient is negative.

Explanation:

(1) Is not the answer because when x = 3, f(x) = 0.

(2) Is not the answer because when f(x) = -3,

x = 0.

(4) Is also not the answer because the y-values on both sides of (1, -4) are mirroring each other.

Therefore,

(3) is the answer, by elimination, but also because there is not proof of the leading coefficient being negative, as we don't see the original equation.

I hope this helped! :)

7 0
3 years ago
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