Tech A says that the difference between stored pulse width and the actual pulse width required to keep the mixture at the correct ratio is called fuel trim. Tech B says that the long-term trim values can be positive or negative. Who is correct?


A. Tech A

B. Tech B

C. Both A and B

D. Neither A nor B

Answers

Answer 1
it has to be C. hope that helps
Answer 2

Tech B says that the long-term trim values can be positive or negative is correct while tech A is wrong.

Why is long term fuel trim negative?

When the displayed fuel trim value is said to be negative number, it implies that the ECU is reducing the injector pulse scope to remove fuel from the air or fuel mixture.

Therefore, Tech B says that the long-term trim values can be positive or negative is correct while tech A is wrong.

So, option B is correct.

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Hi there!

1. C) Ancient Structure

2. A) Iterate

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The first two, I’m sure the third i’m not.

Sorry if I’m wrong and hope this helps !
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Sorry If this is wrong but I’m 90 % sure it’s right. Have a good day!

storage silos can be used to store grains.The silos are 15'wide.The larger silo is 18'10 tall and the shorter is 10'10 tall.NOT INCLUDING THE DOME.how much grain can you store these two silos,combined.Give your answer in cubic feet,to the nearest 0.01.

Answers

Answer:

The amount of rains that can be stored in the two silos combined is 5242.53 ft³

Explanation:

The parameters given are;

Diameter, D, of the silos = 15 feet

Height, h₁, of the larger silo = 18 feet 10 inches = 18.83 feet

Height, h₂, of the shorter silo = 10 feet 10 inches = 10.83 feet

The volume, V, of the cylindrical shape is given by the formula;

Volume = Area of base × Height

[tex]Area \, of \, the \, base = \pi \times \dfrac{D^{2}}{4}[/tex]

[tex]\therefore Volume = \pi \times \dfrac{D^{2}}{4}\times h[/tex]

Therefore, for the larger silo, we have;

[tex]Volume, V_1 = \pi \times \dfrac{D^{2}}{4}\times h_1[/tex]

[tex]V_1 = \pi \times \dfrac{15^{2}}{4}\times 18.83 = 3328.12 \ ft^3[/tex]

for the shorter silo, we have;

[tex]Volume, V_2 = \pi \times \dfrac{D^{2}}{4}\times h_2[/tex]

[tex]V_2 = \pi \times \dfrac{15^{2}}{4}\times 10.83 = 1914.41\ ft^3[/tex]

The amount of rains that can be stored in the two silos combined = 3328.12 ft² + 1914.41 ft² = 5242.53 ft².

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Explanation:

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ANSWER ASAPP!!!! EASY!!

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