50 points and brainliest to correct answer

I need help making a proof for geometry! Here is the question I really need help on this question my marking period ends soon!

Anni [7]

Answer:

Step-by-step explanation:

7 0

3 years ago

Manuel bought 2kg of jelly beans and 2kg of gummy worms for a total of $26. Danny bought 2kg of jelly beans and 3kg of gummy wor

charle [14.2K]

Answer:

Jelly beans: $6 for 1 kg

Gummy worms: $7 for 1 kg

Step-by-step explanation:

In this question, we need to find the cost of 1kg of jelly beans and gummy worms.

To do this, make equations for both scenarios and solve:

2j + 2g = 26

2j + 3g = 33

Turn the bottom equation negative, as we will first solve for gummy worms (g).

2j + 2g = 26

-2j - 3g = -33

Solve:

-g = -7

Divide both sides by -1.

g = 7

We now know that gummy worms cost $7 for 1 kg

To solve for jellybeans (j), plug in 7 to g in one of the equations and solve:

2j + 2(7) = 26

2j + 14 = 26

Subtract 14 from both sides.

2j = 12

Divide both sides by 2.

j = 6

Jellybeans cost $6 for 1 kg.

Check answer by plugging in values to one of the equations:

2(6) + 2(7) = 26

12 + 14 = 26

26 = 26

3 0

3 years ago

If a container contains 18,000 ounces (oz) of beans, about how many pounds (lb) does it contain? [1 pound = 16 ounces] 70 lb 1,1

Margarita [4]

Answer:

The answer is B. 1,125

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

A tank is filled at a constant rate. 10 minutes after filling is started, the tank contains 4.8L of water. After 35 minutes the

Mashcka [7]

Answer:

Part A)

0.1 liters per minute.

Part B)

There was initially 3.8 liters of water.

$\displaystyle V(t) = 0.1(t - 10) + 4.8$

Part C)

562 minutes.

Step-by-step explanation:

A tank is filled at a constant rate. After 10 minutes, the tank contains 4.8 L of water and after 35 minutes, the tank contains 7.3 L of water.

Part A)

We can represent the current data with two points: (10, 4.8) and (35, 7.3). The x-coordinate is measured in minutes since the tank began to be filled and the y-coordinate is measured in how full the tank is in liters.

To find the rate at which the tank is being filled, find the slope between the two points:

$\displaystyle m = \frac{\Delta y}{\Delta x} = \frac{(7.3)-(4.8)}{(35)-(10)} = \frac{2.5}{25} = 0.1$

In other words, the rate at which the tank is being filled is 0.1 liters per minute.

Part B)

To find the function of the volume of the tank, we can use the point-slope form to first find its equation:

$\displaystyle y - y_1 = m( x - x_1)$

Where m is the slope/rate of change and (x₁, y₁) is a point.

We will substitute 0.1 for m and let (10, 4.8) be the point. Hence:

$\displaystyle y - (4.8) = 0.1(x - 10)$

Simplify:

$\displaystyle y = 0.1(x-10) + 4.8$

Since y represent how full the tank is and x represent the time in minutes since the tank began to be filled, we can substitute y for V(t) and x for t. Thus, our function is:

$\displaystyle V(t) = 0.1(t - 10) + 4.8$

The initial volume is when t = 0. Evaluate:

$\displaystyle V(0) = 0.1 ((0) - 10) + 4.8 = 3.8$

There was initially 3.8 liters of water.

Part C)

To find how long it will take for the tank to be completely filled given its maximum capacity of 60 liters, we can let V(t) = 60 and solve for t. Hence:

$60 = 0.1(t - 10) + 4.8$

Subtract:

$55.2 = 0.1(t - 10)$

Divide:

$552 = t - 10$

Add. Therefore:

$t = 562\text{ minutes}$

It will take 562 minutes for the tank to be completely filled.

8 0

3 years ago

How to isolate scalars from a matrix

I am Lyosha [343]

The technique of matrix isolation involves condensing the substance to be studied with a large excess of inert gas (usually argon or nitrogen) at low temperature to form a rigid solid (the matrix). The early development of matrix isolation spectroscopy was directed primarily to the study of unstable molecules and free radicals. The ability to stabilise reactive species by trapping them in a rigid cage, thus inhibiting intermolecular interaction, is an important feature of matrix isolation. The low temperatures (typically 4-20K) also prevent the occurrence of any process with an activation energy of more than a few kJ mol-1. Apart from the stabilisation of reactive species, matrix isolation affords a number of advantages over more conventional spectroscopic techniques. The isolation of monomelic solute molecules in an inert environment reduces intermolecular interactions, resulting in a sharpening of the solute absorption compared with other condensed phases. The effect is, of course, particularly dramatic for substances that engage in hydrogen bonding. Although the technique was developed to inhibit intermolecular interactions, it has also proved of great value in studying these interactions in molecular complexes formed in matrices at higher concentrations than those required for true isolation.

3 0

3 years ago