WILL GIVE BRAINLIEST
WILL GIVE BRAINLIEST - 1

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Answer 1
Answer:

Answere:

its black

Step-by-step explanation:

black is better and im right


Related Questions

A particle moving along a line has position s(t) = t^4 − 20t^2 m at time t seconds. Determine: (a)- At which times does the particle pass through the origin? (b)- At which times is the particle instantaneously motionless.
Question below. no links please!
Write the equation in standard form and then find the center (h, k) andx^2+8x+ y2 + 4y - 5= 0
The tread life of tires mounted on light-duty trucks follows the normal probability distribution with a mean of 60,000 miles and a standard deviation of 4,000 miles. Suppose you bought a set of four tires, what is the likelihood the mean tire life of these four tires is between 57,000 and 63,000 miles
Carlos purchased an antique chair for $56.He later sold the chair for $68 to an antiquedealer. What was the percent markup of thechair?

Use the quadratic formula to solve the equation 2y^ +6y-8=0

Answers

To solve this problem you must apply the proccedure shown below:

 1. You have the following quadratic equation, which is given in the problem above:

 2y^2+6y-8=0

 2. So, to solve it, you can use the quadratic formula, which is:

 y=(-b
±√(b²-4ac))/2a

 a=2
 b=6
 c=-8

 3. When you susbtitute these values into the formula, you obtain:

 y1=-4
 y2=1

At a farm, there are two huge pits for storing hay. 90 tons of hay is stored in the first pit, 75 tons in the second pit. Then, three times as much hay was removed from the first pit as was removed from the second pit. After that, there was half as much hay in the first pit as there was now in the second pit. How many tons of hay was taken from the first pit?

Answers

Answer:

63 tons

Step-by-step explanation:

The problem statement asks for the tons of hay removed from the first pit. It is convenient to let a variable (x) represent that amount. This is said to be 3 times the amount removed from the second pit, so that amount must be x/3.

The amount remaining in the first pit is 90-x.

The amount remaining in the second pit is 75 -x/3.

Since the first pit remaining amount is half the second pit remaining amount, we can write the equation ...

... 90 -x = (1/2)(75 -x/3)

... 180 -2x = 75 -x/3 . . . . multiply by 2

... 105 - 2x = -x/3 . . . . . . subtract 75

... 315 -6x = -x . . . . . . . . multiply by 3

... 315 = 5x . . . . . . . . . . . add 6x

... 63 = x . . . . . . . . . . . . . divide by 5

63 tons of hay were taken from the first pit.

_____

Check

After removing 63 tons from the first pit, there are 27 tons remaining. After removing 63/3 = 21 tons from the second pit, there are 54 tons remaining. 27 is half of 54, so the answer checks OK.

• Which factors do 3 and 9 have in common? Choose ALL that apply.1
2
3
4
5
6
7
8
9
10

Answers

Answer:

3 and 9 both have factors 1 and 3 in common.

Explanation:

A factor is a number that divides into another number exactly and without leaving a remainder. For instance, factors of 15 are 3 and 5, because 3×5 = 15. Some numbers have more than one factorization (more than one way of being factored). For example, 12 can be factored as 1×12, 2×6, or 3×4.

I hope this was helpful to you! If it was, please consider rating, pressing thanks, and marking my answer as 'Brainliest.' Have a wonderful day.

3 and 9 are the only ones that are in common for both 3 and 9

A rectangular block of aluminum30 mm×60 mm×90 mm is placed in apressurechamber and subjected to a pressure of 100 MPa. If the modulus of elasticity is 75GPa andPoisson's ratio is 0.35, what will be the decrease in the longest side of the block, assuming thatthe material remains within the linear elastic region? What will be the decrease in the volume oftheblock? g

Answers

Answer:

\Delta V = - 216.415 mm^3

Step-by-step explanation:

we know that change is length is calculated by following strain relation

\Delta L = L * \epsilon_x

where strain is given as

\epsilon_x = (\sigma_x - \nu(\sigma_y + \sigma_z))/(E)

\epsilon_x = (-10^8 - 0.35 ( -10^8 -10^8 (N/m^2)))/(7.5 * 10^10)

\epsilon_x = -4.453 * 10^(-4)

plugging strain value in change in length formula

\Delta L =  90 *  -4.453 * 10^(-4)  = - 0.04008 mm

calculate the length on the longer side

L_(long) = L = \delta L

              = 90 - 0.04008 = 89.95 mm

intial volume =  90*60*30 = 1.62 * 10^5 mm^3

change in volume

\Delta V =V ( \epsilon_x +\epsilon_y +  \epsilon_z )

\Delta V = 1.62 * 10^5 (-4.453 - 4.453 - 4.453) * 10^(-4)

\Delta V = - 216.415 mm^3

Final answer:

Calculations involve determining strain from given pressure and Modulus of Elasticity and then determining the decrease in length of the longest side and total volume of the aluminum block when subjected to pressure.

Explanation:

The question is about applying principles of material science under conditions of pressure. The decrease in length and volume of a rectangular block of aluminum when subjected to pressure can be calculated by using the concepts of Modulus of Elasticity and Poisson's Ratio.

First, the strain experienced can be calculated using the formula:

Strain = Pressure / Modulus of Elasticity

Substituting the given values, the strain is found. The change in the longest side (90mm) is calculated by multiplying the original length by the strain. The volume change is calculated using the formula:

Change in volume = Original volume * (-3) * strain

Where original volume is = 30mm * 60mm * 90mm. Here the negative indicates a decrease. This will provide the decrease in the longest side and the total volume of the block when subjected to the given pressure.

Learn more about Material Deformation under Pressure here:

brainly.com/question/31787736

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Drag each question to the correct location on the table. Classify the questions as statistical or not statistical.

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Answer:

wheres the picture

Step-by-step explanation:

Answer:

picture?

Step-by-step explanation:

I need help with this question

Answers

Answer:
_______________________________________________
1)  72\pi cm² / min.  ;
_______________________________________________
2)  192 \pi cm² / min.  
_______________________________________________
Explanation:
____________________________________________
     
 Area of a circle:  A = \pi r² ;

Rate of change of Area is:  da/dt = 2\pi r (dr/dt) ;

Given:  dr/dt = 3 cm / min.
______________________________________________________
 Problem 1)
____________________________________________________________
dA/dt = 2 \pi *(12 cm)*(3 cm/ min) = 72 \pi  cm² / min. ;
____________________________________________________________
 Problem 2)
____________________________________________________________
dA/dt = 2 \pi *(32 cm)*(3 cm/ min) = 192 \pi  cm² / min.
_____________________________________________________________