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natulia [17]
3 years ago
15

Technician A says when there is a delay or slip as the transmission shifts into any forward gear, a slipping rear clutch is indi

cated. Technician B says a slipping front or forward clutch is indicated if the shift for all forward gears is delayed. Who is correct?
Mathematics
1 answer:
blagie [28]3 years ago
8 0

Answer:

Technician B

Step-by-step explanation:

It is very important that any car is taking for a test drive and a road test is carried out to check for any sort of problem with the transmission of the car or any leakages that may be found in the hydraulic circuits.

Technician B is very correct because when the hydraulic circuits is leaking, the front or forward clutch tends to slip thereby causing the shift for all forward gears to be delayed

Also,the sticking of the spool into the body of the valve can also cause the problem to occur as well.

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3 0
3 years ago
The following stock prices reflect the closing price for the past five days. Find the 3 day SMA
Masteriza [31]

Answer:

solution

Step-by-step explanation: total=209.75

7 0
2 years ago
Can someone help me with this, pleasee.
Schach [20]

Answer:

i will say the first one cannot, the second one cannot and the third one can

Step-by-step explanation:

but i am not 100% sure sry

5 0
3 years ago
Read 2 more answers
Someone help me with this
Jet001 [13]

2a. 14,9; 2b. 15,4; 2c. 30,8; 3. 32

For 2a., you have to set it up like this: csc 48° = 20⁄x [OR sin 48° = ˣ⁄20]. Then you would have to isolate the variable by getting rid of the denominator. The cosecant function has an extra step because you will get xcsc 48° = 20. As stated, isolate the variable; this time, divide by csc 48°. This is what 20 will be divided by to get your x, whereas the other side cancels out. Then you have to round to the nearest tenth degree.

For 2b., you have to set it up like this: sec 39° = ˣ⁄12 [OR cos 39° = 12⁄x. Then you would have to isolate the variable by getting rid of the denominator. The cosine function has an extra step because you will get xcos 39° = 12. As stated, isolate the variable; this time, divide by cos 39°. This is what 12 will be divided by to get your x, whereas the other side cancels out. Then you have to round to the nearest tenth degree.

For 2c., you have to set it up like this: cot 64° = 15⁄x [OR tan 64° = ˣ⁄15]. Then you would have to isolate the variable by getting rid of the denominator. The cotangent function has an extra step because you will get xcot 64° = 15. As stated, isolate the variable; this time, divide by cot 64°. This is what 15 will be divided by to get your x, whereas the other side cancels out. Then you have to round to the nearest tenth degree.

Now, for 3., it is unique, but similar concept. In this exercise, we are solving for an angle measure, so we have to use inverse trigonometric ratios. So, we set it up like this: cot⁻¹ 1⅗ = m<x [OR tan⁻¹ ⅝ = m<x]. We simply input this into our calculator and we get 32,00538321°. When rounded to the nearest degree, we get 32°.

WARNING: If you use a graphing calculator, you have to input it uniquely because most graphing calculators do not have the inverse trigonometric ratios programmed in their systems. This is how you would write this: tan⁻¹ 1⅗⁻¹. You set 1⅗ to the negative first power, ALONG WITH the inverse tangent function, because without it, your answer will be thrown off. Since Cotangent and Tangent are multiplicative inverses of each other, that is the reason why the negative first power is applied ALONG WITH the inverse tangent function.

**NOTE: 1⅗ = 8⁄5

Take into consideration:

sin <em>θ</em> = O\H

cos <em>θ</em><em> </em>=<em> </em>A\H

tan <em>θ</em><em> </em>= O\A

sec <em>θ</em><em> </em>= H\A

csc <em>θ</em><em> </em>= H\O

cot <em>θ</em><em> </em>= A\O

I hope this helps you out alot, but if you are still in need of assistance, do not hesitate to let me know and subscribe to my channel [username: MATHEMATICS WIZARD], and as always, I am joyous to assist anyone at any time.

7 0
3 years ago
Joni has two cats in the yard, Stephen and Graham. If both animals gained 20 pounds, Stephen would be 25% heavier than Graham. F
tresset_1 [31]

Answer:

Stephen weighs 80 lbs

Graham weighs 60 lbs

Step-by-step explanation:

Let Stephen the cat = S

Let Graham the cat = G

25% heavier = 125% = 125/100 = 1.25

50% heavier = 150% = 150/100 = 1.5

If both animals gained 20 lbs, S would be 25% heavier than G:

    S + 20 = 1.25(G + 20)  

⇒          S = 1.25(G + 20) - 20

⇒          S = 1.25G + 5

If both animals lost 20 lbs, S would be 50% heavier than G:

    S - 20 = 1.5(G - 20)  

⇒          S = 1.5(G - 20) + 20

⇒          S = 1.5G - 10

Equate both equations for S, and solve for G:

                 S = S

⇒ 1.25G + 5 = 1.5G - 10

⇒     -0.25G = -15

⇒              G = 60

Substitute found value for G into one of the equations for S and solve for S:

    S = 1.5G - 10  

⇒  S = 1.5 x 60 - 10

⇒  S = 90 - 10

⇒  S = 80

Stephen weighs 80 lbs

Graham weighs 60 lbs

6 0
2 years ago
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