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kondor19780726 [428]
3 years ago
6

Select the system of linear inequalities whose solution is graphed.

Mathematics
1 answer:
Volgvan3 years ago
8 0
The second one is your answer 
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Find all the second partial derivatives. v = e5xey
Helga [31]
I'm taking the liberty of editing your  function  <span>v = e5xey:  It should be 
</span>
<span>v = e^5x^ey, with " ^ " indicating exponentiation.
</span>
Did you mean e^(5x) or (e^5)x?  I'll assume it's e^(5x).

The partial of   v = e^(5x)e^y with respect to x is e^(5x)(5)*e^y, or 25x*e^y.

The partial of v = e^(5x)e^y with respect to y is e^(5x)e^y.
5 0
3 years ago
Can you solve the equation also? ( make sure to add the answer )
polet [3.4K]

Answer:

35-22=x

35-22=13

Step-by-step explanation:

Include units if necessary:

$35-$22=$13

8 0
2 years ago
Read 2 more answers
For which x is f(x)?=-3
Rzqust [24]

Answer:

the answer would be x=10

Step-by-step explanation:

cuz you do a t-chart the put x/-3 then + 10 then - witch is

x=10

8 0
3 years ago
John is at a local bait shop; he wants to buy bait for his fishing trip. At the store, they are selling live bait for $12 a poun
galina1969 [7]

Answer:

x\geq3

12x+7y\leq63

y\geq0

Step-by-step explanation:

If we let x be the amount of live bait and y be the amount of natural bait, Then we can come up with the following inequalities;

We are told that John would like to get at least 3 pounds of live bait. At least 3 means 3 or more. Since x represents the amount of live bait, we have;

x\geq3

Moreover,we are informed that;

The store sells live bait for $12 a pound and natural bait for $7 a pound. x pounds of live bait would cost 12x while y pounds of natural bait would cost 7y. The total cost would thus be;

12x + 7y

but John only has a budget of $63. This implies that he can spend $63 at most, thus;

12x+7y\leq63

Finally we can have our last inequality as;

y\geq0

8 0
3 years ago
SHOW YOUR WORK,Solve this system of equations for x using the addition method
maria [59]

Answer:

NUMBER 1.)

Step 1

Subtract 3y3y from both sides.

5x=10-3y5x=10−3y

Step 2

Divide both sides by 55.

\frac{5x}{5}=\frac{10-3y}{5}

5

5x

=

5

10−3y

Hint

Undo multiplication by dividing both sides by one factor.

Step 3

Dividing by 55 undoes the multiplication by 55.

x=\frac{10-3y}{5}x=

5

10−3y

Hint

Undo multiplication.

Step 4

Divide 10-3y10−3y by 55.

x=-\frac{3y}{5}+2x=−

5

3y

+2

Hint

Divide.

Solution

x=-\frac{3y}{5}+2x=−5

3y+2

Step-by-step explanation:

NUMBER 2.)

Step 1

Add 4y4y to both sides.

3x=6+4y3x=6+4y

Step 2

The equation is in standard form.

3x=4y+63x=4y+6

Step 3

Divide both sides by 33.

\frac{3x}{3}=\frac{4y+6}{3}

3

3x

=

3

4y+6

Hint

Undo multiplication by dividing both sides by one factor.

Step 4

Dividing by 33 undoes the multiplication by 33.

x=\frac{4y+6}{3}x=

3

4y+6

Hint

Undo multiplication.

Step 5

Divide 6+4y6+4y by 33.

x=\frac{4y}{3}+2x=

3

4y

+2

Hint

Divide.

Solution

x=\frac{4y}{3}+2x= 3

4y+2

4 0
2 years ago
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