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3 years ago

Find angle M and angle J of the isosceles trapezoid

Mathematics

1 answer:

fgiga [73]3 years ago

3 0

Angle J would be 118 degrees and angle M would be 62 degrees. this is because it’s an isosceles trapezoid and corresponding angles are the same.

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–2x2 – 12x – 9 = 0 PLS HELP I DO NOT WANT TO FAIL I WILL GIVE BRAINLIEST

Pie

The right answer is -13/12.

4 0

3 years ago

Define Following:consistent equations

mario62 [17]

Answer:

1. Consistent equations

x + y = 3

x + 2·y = 5

2. Dependent equations

3·x + 2·y = 6

6·x + 4·y = 12

3. Equivalent equations

9·x - 12·y = 6

3·x - 4·y = 2

4. Inconsistent equations

x + 2 = 4 and x + 2 = 6

5. Independent equations

y = -8·x + 4

8·x + 4·y = 0

6. No solution

4 = 2

7. One solution

3·x + 5 = 11

x = 2

Step-by-step explanation:

1. Consistent equations

A consistent equation is one that has a solution, that is there exist a complete set of solution of the unknown values that resolves all the equations in the system.

x + y = 3

x + 2·y = 5

2. Dependent equations

A dependent system of equations consist of the equation of a line presented in two alternate forms, leading to the existence of an infinite number of solutions.

3·x + 2·y = 6

6·x + 4·y = 12

3. Equivalent equations

These are equations with the same roots or solution

e.g. 9·x - 12·y = 6

3·x - 4·y = 2

4. Inconsistent equations

Inconsistent equations are equations that are not solvable based on the provided set of values in the equations

e.g. x + 2 = 4 and x + 2 = 6

5. Independent equations

An independent equation is an equation within a system of equation, that is not derivable based on the other equations

y = -8·x + 4

8·x + 4·y = 0

6. No solution

No solution indicates that the solution is not in existence

Example, 4 = 2

7. One solution

This is an equation that has exactly one solution

Example 3·x + 5 = 11

x = 2

3 0

4 years ago

Alex is trying to run a certain number of miles by the end of the month Alex is 40% of the way to achieving her goal and she alr

docker41 [41]

Answer:

30 miles

Step-by-step explanation:

Given that:

Alex has some target to run a certain number of miles by the end of the month.

Goal already achieved = 40% of the total goal

Number of miles already run by Alex = 12 miles

To find:

Number of miles that Alex is trying to run by the end of month?

Solution:

We have to find nothing but the goal of Alex here.

Let the number of miles that Alex is trying to run by the end of the month = $x$ miles

As per question statement:

40% of total number of miles to be run = 12 miles

$\dfrac{40}{100}\times x = 12\\\Rightarrow 40x=12\times 100\\\Rightarrow x = \dfrac{1200}{40}\\\Rightarrow \bold{x = 30\ miles}$

Total number of miles that Alex is trying to run by the end of the month = <em>30 miles</em>

4 0

3 years ago

Suppose brine containing 0.2 kg of salt per liter runs into a tank initially filled with 500 L of water containing 5 kg of salt.

Oliga [24]

Answer:

(a) 0.288 kg/liter

(b) 0.061408 kg/liter

Step-by-step explanation:

(a) The mass of salt entering the tank per minute, x = 0.2 kg/L × 5 L/minute = 1 kg/minute

The mass of salt exiting the tank per minute = 5 × (5 + x)/500

The increase per minute, Δ/dt, in the mass of salt in the tank is given as follows;

Δ/dt = x - 5 × (5 + x)/500

The increase, in mass, Δ, after an increase in time, dt, is therefore;

Δ = (x - 5 × (5 + x)/500)·dt

Integrating with a graphing calculator, with limits 0, 10, gives;

Δ = (99·x - 5)/10

Substituting x = 1 gives

(99 × 1 - 5)/10 = 9.4 kg

The concentration of the salt and water in the tank after 10 minutes = (Initial mass of salt in the tank + Increase in the mass of the salt in the tank)/(Volume of the tank)

∴ The concentration of the salt and water in the tank after 10 minutes = (5 + 9.4)/500 = (14.4)/500 = 0.288

The concentration of the salt and water in the tank after 10 minutes = 0.288 kg/liter

(b) With the added leak, we now have;

Δ/dt = x - 6 × (14.4 + x)/500

Δ = x - 6 × (14.4 + x)/500·dt

Integrating with a graphing calculator, with limits 0, 20, gives;

Δ = 19.76·x -3.456 = 16.304

Where x = 1

The increase in mass after an increase in = 16.304 kg

The total mass = 16.304 + 14.4 = 30.704 kg

The concentration of the salt in the tank then becomes;

Concentration = 30.704/500 = 0.061408 kg/liter.

6 0

3 years ago

Can someone help me with this i am confused

Eva8 [605]

Plot each point on a graph , then count how many you need to go up and then over in this case it is 8 over 1 then calculate the y int so y= 8x -25

6 0

3 years ago