Solution:
1. Consider two trees which are 5 m apart. Suppose one tree is located at a distance of 45 m from my house and another tree is located 50 m from my house. Represent , distance of two trees from my house in terms of Quadratic function.
Solution:
Let location of my house when map of planet earth is drawn or on globe=x
⇒(x-45)(x-50)=0
⇒ x² - 45 x - 50 x +45 × 50 =0
⇒ x² - 95 x + 2250=0
2. Suppose age of me and my brother is 6 and 10 years. The Product of difference between age of me and my mother who is x years old and my brother and mother is 780.Find my mother's age.
Solution: Age of my mother = x years
→(x-6) × (x-10)= 780,
if you will solve it , the value of x= 36 years
The distance covered in one hour is 76 miles.
<u>Step-by-step explanation:</u>
Edward drives a car for 28.5 miles.
Therefore, the distance he travels by the car = 28.5 miles.
The time it takes to drive 28.5 miles is 3/8 of an hour.
<u>The formula to find the rate :</u>
Rate = Distance traveled / Time taken in hours.
Here, the distance is 28.5 miles and time is 3/8 hour.
Rate = 
<u>To find the distance covered in 1 hour :</u>
Let us assume, 'x' be the distance covered in one hour.
Rate = 
<u>Now, comparing the rates to find the x value :</u>
⇒ 
⇒ 
⇒ 
⇒ 
Therefore, the distance covered in one hour is 76 miles.
Answer:
Step-by-step explanation:
this is simple math ur in like second grade,figure this out yourself stop cheating.
Answer: For the sum of 130
First: $90
Second: $40
Step-by-step explanation:
We write equations for each part of this situation.
<u>The Total Charge</u>
Together they charged 1550. This means 1550 is made up of the first mechanics rate for 15 hours and the second's rate for 5 hours. Lets call the first's rate a, so he charges 15a. The second's let's call b. He charges 5b. We add them together 15a+5b=1550.
<u>The Sum of the Rates</u>
Since the first's rate is a and the second is b, we can write a+b=130 since their sum is 130.
We solve for a and b by substituting one equation into another. Solve for the variable. Then substitute the value into the equation to find the other variable.
For a+b=130, rearrange to b=130-a and substitute into 15a+5b=1550.
15a + 5 (130-a)=1550
15a+650-5a=1550
10a+650-650=1550-650
10a=900
a=$90 was charged by the first mechanic.
We substitute to find the second mechanic's rate.
90+b=130
90-90+b=130-90
b= $40 was charged by the second mechanic
Answer:
B
Step-by-step explanation:
