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2 years ago

12

Joe goes to the store with $15 to buy salad. Bags of pre-washed salad on sale for $2 each. What are the possible numbers of bags

joe can buy?

1 answer:

MA_775_DIABLO [31]2 years ago

3 0

Answer:

7

Step-by-step explanation:

15/2= 7.5

he can't buy a half bag so it's 7 bags

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In a small community, a survey on the favorite television programs of 63 housewives was made. Based on the survey, 38 housewives

scoundrel [369]

Answer:

x = 4

Step-by-step explanation:

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2 years ago

The gold bar has a trapezium cross-sectional area. Gold has a density of 19.3 grams per cm3. Work out the mass of the gold bar.

Varvara68 [4.7K]

Answer:

Mass of gold bar = 13.896 kg

Step-by-step explanation:

Volume of gold bar = Area of trapezium x Length of bar

Length of bar = 16 cm

Area of trapezium = 2 x Area of triangle + Area of rectangle

Area of triangle = 0.5 x 5 x 3 = 7.5 cm²

2 x Area of triangle = 2 x 7.5 = 15 cm²

Area of rectangle = 6 x 5 = 30 cm²

Area of trapezium = 15 + 30 = 45 cm²

Volume of gold bar = 45 x 16 = 720 cm³

Density of gold = 19.3 g/cm³

Mass of gold bar = 720 x 19.3 = 13896 g = 13.896 kg

Mass of gold bar = 13.896 kg

6 0

2 years ago

Read 2 more answers

The quotient of the sum of 2t and 2, twice the cube of s

natali 33 [55]

Answer: $\frac{2t+2}{2s^3}$

Step-by-step explanation:

Since you did not indicate what you need to do, I assume that you have to write an expression using the sentence given in the problem.

In order to solve this exercise, it is importat to remember the following information:

1. The quotient is the result of a division.

2. The sum is the result of an addition.

3. The word "twice" indicates a multiplicatio by 2.

4. The word "cube" indicates an exponent 3.

Then, keeping on mind the explained above and the data given in the exercise, you know that:

-The sum of $2t$ and $2$ can be expressed as:

$2t+2$

- Twice the cube of $s$ can be expressed in the following form:

$2s^3$

Therefore, you can dermine that "the quotient of the sum of $2t$ and $2$ and twice the cube of $s$ " is represented with the following expression:

$\frac{2t+2}{2s^3}$

O

8 0

2 years ago

Tommy posts 11 pictures on Instagram every day . Elizabeth posts 15 pictures on Instagram every day .

sleet_krkn [62]

You can just think of any variable for example let's use "T" , "E". Then do T +E for part a. Then do T=11 E=15. After do 11 •7 and 15•7. 11•7=77 . 15•7=105. Last do 105+77 and it equals 182.

7 0

2 years ago

Read 2 more answers

Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint. (If an answer d

andrey2020 [161]

Answer:

The maximum value is 1/27 and the minimum value is 0.

Step-by-step explanation:

Note that the given function is equal to $(xyz)^2$ then it means that it is positive i.e $f(x,y,z)\geq 0$ .

Consider the function $F(x,y,z,\lambda)=x^2y^2z^2-\lambda (x^2+y^2+z^2-1)$

We want that the gradient of this function is equal to zero. That is (the calculations in between are omitted)

$\frac{\partial F}{\partial x} = 2x(y^2z^2 - \lambda)=0$

$\frac{\partial F}{\partial y} = 2y(x^2z^2 - \lambda)=0$

$\frac{\partial F}{\partial z} = 2z(x^2y^2 - \lambda)=0$

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

Note that the last equation is our restriction. The restriction guarantees us that at least one of the variables is non-zero. We've got 3 options, either 1, 2 or none of them are zero.

If any of them is zero, we have that the value of the original function is 0. We just need to check that there exists a value for lambda.

Suppose that x is zero. Then, from the second and third equation we have that

$-2y\lambda = -2z\lambda$ . If lambda is not zero, then y =z. But, since $-2y\lambda=0$ and lambda is not zero, this implies that x=y=z=0 which is not possible. This proofs that if one of the variables is 0, then lambda is zero. So, having one or two variables equal to zero are feasible solutions for the problem.

Suppose that only x is zero, then we have the solution set $y^2+z^2=1$ .

If both x,y are zero, then we have the solution set $z^2=1$ . We can find the different solution sets by choosing the variables that are set to zero.

NOw, suppose that none of the variables are zero.

From the first and second equation we have that

$\lambda = y^2z^2 = x^2z^2$ which implies $x^2=y^2$

Also, from the first and third equation we have that

$\lambda = y^2z^2 = x^2y^2$ which implies $x^2=z^2$

So, in this case, replacing this in the restriction we have $3z^2=1$ , which gives as another solution set. On this set, we have $x^2=y^2=z^2=\frac{1}{3}$ . Over this solution set, we have that the value of our function is $\frac{1}{3^3}= \frac{1}{27}$

4 0

2 years ago