Answer:
Step-by-step explanation:
1: SA = bh + (s1 + s2 + s3)H
2: A = lw
3: A = lw
Answer:
hi there ☺️
Here we will use algebra to find three consecutive integers whose sum is 345. We start by assigning X to the first integer. Since they are consecutive, it means that the 2nd number will be X + 1 and the 3rd number will be X + 2 and they should all add up to 345. Therefore, you can write the equation as follows:
(X) + (X + 1) + (X + 2) = 345
To solve for X, you first add the integers together and the X variables together. Then you subtract three from each side, followed by dividing by 3 on each side. Here is the work to show our math:
X + X + 1 + X + 2 = 345
3X + 3 = 345
3X + 3 - 3 = 345 - 3
3X = 342
3X/3 = 342/3
X = 114
Which means that the first number is 114, the second number is 114 + 1 and the third number is 114 + 2. Therefore, three consecutive integers that add up to 345 are 114, 115, and 116.
114 + 115 + 116 = 345
We know our answer is correct because 114 + 115 + 116 equals 345 as displayed above.
Step-by-step explanation:
pls rate me the brainiest
Answer:
I KNOWW THE ANSWER!!
Step-by-step explanation:
Chapter : Algebra
Study : Math in Junior high school
x = 7 + √40
find √x of √x + 1
= √x + 1
= √(7+√40) + 1
in Formula is :
= √7+√40 = √x + √y
= (√7+√40)² = (√x + √y)²
= 7+√40 = x + 2√xy + y
= 7 + √40 = x + y + 2√xy
→ 7 = x + y → y = 7 - x ... Equation 1
→ √40 = 2√xy → √40 = 2.2√10 = 4√10
= xy = 10 ... Equation 2
substitution Equation 1 to 2 :
= xy = 10
= x(7-x) = 10
= 7x - x² = 10
= x² - 7x + 10 = 0
= (x - 5)(x - 2) = 0
= x = 5 or x = 2
Subsitution x = 5 and x = 2, to equation 1
#For x = 5
= y = 7 - x
= y = 7 - (5)
= y = 2
#For x = 2
= y = 7 - x
= y = 7 - (2)
= y = 5
and his x and y was find :
#Equation 1 :
= x = 5 and y = 2
#Equation 2 :
= x = 2 and y = 5
So that :
√7+√40 = √x + √y
= √7+√40 = √2 + √5
And that is answer of question :
= √2 + √5 + 1
To solve this I'm going to split the middle term.
First multiply the first and last terms:
24x^2
So find two numbers that multiply to 24x^2 and add to 11x.
This would be 3x and 8x
Rewrite the problem as
4x^2+3x+8x+6
Take the first and 3rd and 2nd and 4th terms
4x^2 and 8x
and
3x and 6
Factor by grouping
Take out a 4x for the first group to get 4x(x+2)
Take out a 3 for the 2nd group to get 3(x+2)
Rewrite as (4x+3)(x+2)
Hope this helps.
