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ella [17]
3 years ago
5

How to solve the problem

Mathematics
1 answer:
luda_lava [24]3 years ago
7 0
Substitue y with the first equation.
x+3=2x+5 (Subtract x on the left sides from both sides.)
3=x+5 (Subtract 5 from both sides.)
x=-2
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5x (x - 4) = 0<br> Solve it in standard form
murzikaleks [220]

Answer:

x=0 or x=4

Step-by-step explanation:

5x(x−4)=0

4 0
3 years ago
3. In a single growing season at the Smith Family Orchard, the average yield per apple tree is 150 apples when the number of tre
g100num [7]

Answer:

<em><u>A.10000</u></em>

<em><u>B.25 more trees must be planted</u></em>

Step-by-step explanation:

⇒Given:

  • The intial average yield per acre y_{i} = 150
  • The initial number of trees per acre  t_{i} = 100
  • For each additional tree over 100, the average yield per tree decreases by 1 i.e , if the number trees become 101 , the avg yield becomes 149.
  • Total yield = (number of trees per acre)*(average yield per acre)

<em>A.</em>

⇒If the total trees per acre is doubled , which means :

total number of trees per acre t_{f} = 2*t_{i} = 200

the yield will decrease by : t_{f} - y_{i}

y_{f}= 150-100= 50

⇒total yield = 50*200=10000

<em>B.</em>

⇒to maximize the yield ,

let's take the number of trees per acre to be 100+y ;

and thus the average yield per acre = 150 - y;

total yield = (100+y)*(150-y)\\=15000+50y-y^{2} \\

this is a quadratic equation. this can be rewritten as ,

     ⇒   =15000+50y-y^{2}\\=15000+625 - (625 - 50y +y^{2})\\=15625 - (y-25)^{2}

In this equation , the total yield becomes maximum when y=25;

<u><em>⇒Thus the total number of trees per acre = 100+25 =125;</em></u>

                 

3 0
3 years ago
Expand or factor each of the following expressions to determine which expressions are equivalent.​
bearhunter [10]
(3x-4y)^2=9x^2-24x+16y^2
(3x-2)(9x^2+6x+4)=27x^3-8
4 0
3 years ago
Read 2 more answers
Can someone please help me
IrinaK [193]
For the mean, add together all of the numbers together (let's call the total x) and count how many numbers there are. (let's call this number y). divide x by y, and that is your mean. for the median, put all the numbers in order, and the median is the number in the middle. if there are two numbers in the middle, add them together and divide them by 2.
6 0
3 years ago
The earth rotates once per day about an axis passing through the north and south poles, an axis that is perpendicular to the pla
forsale [732]

Answer:

a)

Speed at Equator = 463.97 meters per second

Centripetal Acceleration at Equator = 3.37*10^{-2} meters per second squared

b)

Speed at 30 degrees north of equator = 401.79 meters per second

Centripetal Acceleration at 30 degrees north of equator = 2.92*10^{-2} meters per second squared

Step-by-step explanation:

The formula is:

v=\frac{2 \pi R}{T}

Where

v is speed

R is radius

T is time

and another formula for centripetal acceleration:

a_c=\frac{4 \pi^{2} R}{T^2}

Now,

a)

at equator, the radius is radius of earth (given), time in seconds is

T = 24 * 60 * 60 = 86,400

THus,

v_E=\frac{2 \pi (6.38*10^{6}}{86,400}=463.97

Speed at Equator = 463.97 meters per second

Centripetal Acceleration:

a_{cE}=\frac{v_E^2}{R_E}=\frac{463.97}{6.38*10^{6}}=3.37*10^{-2}

Centripetal Acceleration at Equator = 3.37*10^{-2} meters per second squared

b)

At 30.0° north of the equator:

R_N=R_E Cos (30)= (6.38*10^6)Cos(30)=5.53*10^6

Now,

Speed = v_{30N}=\frac{2 \pi (5.53*10^6)}{86,400}=401.79

Speed at 30 degrees north of equator = 401.79 meters per second

Centripetal Acceleration:

a_{30N}=\frac{v_E^2}{R_E}=\frac{401.79}{5.53*10^6}=2.92*10^{-2}

Centripetal Acceleration at 30 degrees north of equator = 2.92*10^{-2} meters per second squared

4 0
3 years ago
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