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3 years ago

Simplify the expression 3√2 - √2

Mathematics

2 answers:

Brilliant_brown [7]3 years ago

8 0

This is equal to 2(sqrt)2

WINSTONCH [101]3 years ago

6 0

2 square root 2

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katelyn bought 5.68 pounds of peaches. the peaches were on sale for 1.49 a pound. how much did lately spend on peaches

Leya [2.2K]

So Katelyn bought 5.68 pounds of peaches. And one pound of peaches cost $1.49. So that would mean that we would have to multiply the pounds of peaches she bought, with how much each pound costs. 5.68 × 1.49 = ?

Once you multiply these two decimals, you would get 8.4632. So she spent $8.4632 on 5.68 pounds of peaches. You would multiply the two decimals to find how much she spent in all.

7 0

3 years ago

Read 2 more answers

What is the length of the side of a right triangle that has a side length of 7 cm and a hypotenuse that measures 25 cm?

Elenna [48]

It would be: 25² = 7² + b²
b² = 625 - 49
b = √576
b = 24

In short, Your Answer would be Option B

Hope this helps!

7 0

3 years ago

The variables y and x have a proportional relationship, and y = 28 when x = 21.

Pie

Answer:

x=35

Step-by-step explanation:

i hope this helps

8 0

3 years ago

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What’s the difference for this equation!!!

Katarina [22]

–(x^2–2x+12) would become –x^2+2x–12.

3x^2+9x+13 – x^2+2x–12

= 2x^2 + 11x – 1

7 0

3 years ago

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This extreme value problem has a solution with both a maximum value and a minimum value. Use Lagrange multipliers to find the ex

azamat

Answer:

Maximum at points (8,0),(-8,0).Minimum at points (0,8), (0,-8).

Step-by-step explanation:

There are multiple ways of using lagrange multipliers. Most of them are equivalent.

Consider the function $F(x,y) = x^2-y^2-\lambda(x^2+y^2-64)$ . We want the following $\frac{\partial F}{\partial x} = \frac{\partial F}{\partial y} = \frac{\partial F}{\partial \lambda} = 0$ .

Then, we have

$\frac{\partial F}{\partial x} = 2x-2x\lambda= 2x(1-\lambda)=0$

$\frac{\partial F}{\partial y} = -2y-2y\lambda = -2y(1+\lambda)=0$

$\frac{\partial F}{\partial \lambda} = x^2+y^2-64=0$

From the first two equations, we can see that if $\lambda =1$ then necessarily y=0. IN that case, from the third equation (which is the restriction) gives us that $x=\pm 8$ .

On the other hand, if $\lambda=-1$ then necessarily x=0. Again, using the restriction this gives us that $y=\pm 8$ .

if we evaluate the original function in this points, we have that $f(0,\pm 8) = -64, f(\pm 8,0)=64$ . Then, we have Maximum at points (8,0),(-8,0) and Minimum at points (0,8), (0,-8).

3 0

4 years ago