--Answer--
The second one, fourth, and fifth.
--Explanation--
1. - 6.283...
2. 7.142...
3. 6.928...
4. 7.483...
5. 7.874...
X/5 when x=30 is 30/5 which is 6
3y^2 when y=2 is 3(2)^2 which is 3(4) which is 12
And z=-2... if we plug these into the original question we get 6+12-(-2)
The negatives cancel and it becomes 6+12+2 which =20
Your answer is 20
They need to put 14.4 grams of onion powder in each bottle.
Given,
The quantity of onion powder in a seasoning blend = 4%
We have to find the quantity of pure onion powder should they include in a 72g bottle to make the final blend have 20% onion powder;
How much is a percentage?
A percentage is a certain number or part in every hundred. It is a fraction with the denominator 100, and the symbol for it is "%."
Here,
Let x be the seasoning blend.
Now, 4% of 72=20% of x
⇒ 0.04×72=0.2×x
⇒ 0.2x=2.88
⇒ x=2.88/0.2
⇒ x=14.4 g
Therefore, a container should contain 14.4 grams of onion powder.
Learn more about quantity of onion powder here;
brainly.com/question/29332582
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Answer:
B) Circle
Step-by-step explanation:
Answer:
10 hours at job (1)
2 hours at job (2)
Step-by-step explanation:
As per the given information, one earns ($8) dollars at one of their jobs, and ($10) hours at the other. One must earn a total of ($100) dollars, and can work no more than (12) hours. Let (x) be the hours worked at job 1 and (y) be the hours worked at job two.
Since one can work no more than (12) hours, the sum of (x) and (y) must be (12), therefore the following equation can be formed;
![x+y=12](https://tex.z-dn.net/?f=x%2By%3D12)
One earns ($8) dollars at one of their jobs and ($10) at the other, but one earns a total of (100) one can form an equation to represent this situation. Multiply the hours worked by the money earn per hour for each job, add up the result and set it equal to (100).
![8x+10y=100](https://tex.z-dn.net/?f=8x%2B10y%3D100)
Now set up these equations in a system;
![\left \{ {{x+y=12} \atop {8x+10y=100}} \right.](https://tex.z-dn.net/?f=%5Cleft%20%5C%7B%20%7B%7Bx%2By%3D12%7D%20%5Catop%20%7B8x%2B10y%3D100%7D%7D%20%5Cright.)
Use the process of elimination to solve this system. The process of elimination is a method of solving a system of equations. One must first manipulate one of the equations in the system such that one of the variable coefficients is the additive inverse of the other. That way, when one adds the equation, the variable cancels, one can solve for the other variable then back solve to find the value of the first variable,
![\left \{ {{x+y=12} \atop {8x+10y=100}} \right.](https://tex.z-dn.net/?f=%5Cleft%20%5C%7B%20%7B%7Bx%2By%3D12%7D%20%5Catop%20%7B8x%2B10y%3D100%7D%7D%20%5Cright.)
Manipulate,
![= \left \{ {{(*-8)(x+y=12)} \atop {8x+10y=100}} \right.\\\\](https://tex.z-dn.net/?f=%3D%20%5Cleft%20%5C%7B%20%7B%7B%28%2A-8%29%28x%2By%3D12%29%7D%20%5Catop%20%7B8x%2B10y%3D100%7D%7D%20%5Cright.%5C%5C%5C%5C)
Simplify,
![= \left \{ {{-8x-8y=-96} \atop {8x+10y=100}} \right.\\](https://tex.z-dn.net/?f=%3D%20%5Cleft%20%5C%7B%20%7B%7B-8x-8y%3D-96%7D%20%5Catop%20%7B8x%2B10y%3D100%7D%7D%20%5Cright.%5C%5C)
Add,
![=2y=4](https://tex.z-dn.net/?f=%3D2y%3D4)
Inverse operations,
![y=2](https://tex.z-dn.net/?f=y%3D2)
Backsolve for (x), use equation one to achieve this,
![x+y=12\\](https://tex.z-dn.net/?f=x%2By%3D12%5C%5C)
Substitute,
![x+2=12](https://tex.z-dn.net/?f=x%2B2%3D12)
Inverse operations,
![x=10](https://tex.z-dn.net/?f=x%3D10)