Remember: PEMDAS - Parentheses, exponents, multiply/divide, add/subtract
Distribute: 9(1+x) = 9+9x
19 - 5x = 9 + 9x
Add 5x to both sides of the equation and subtract 9 from both sides.
19 - 5x + (5x) = 9 + 9x + (5x)
19 - (9) = 9 - (9) + 14x
10 = 14x
Divide both sides by 14 to get the variable on its own.
10/14 = 14x/14
10/14=x
5/7 = x
The least common multiple of each pair of the polynomial (5y² - 80) and
(y + 4) is equal to 5(y-4)(y+4).
As given in the question,
Given pair of the polynomial is (5y² - 80) and (y + 4)
Simplify the given polynomial using (a² -b²) = (a-b)(a +b)
(5y² - 80) = 5(y² -16)
⇒(5y² - 80) = 5(y² - 4²)
⇒(5y² - 80) = 5(y -4)(y + 4)
And (y + 4) = (1) (y+4)
Least common multiple = 5(y-4)(y+ 4)
Therefore, the least common multiple of the given pair of the polynomial is 5(y -4)(y+ 4).
Learn more about least common multiple here
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<span>-1 + n=8(n+6)
Use distributive property
-1 + n= 8n + 48
Subtract n from both sides
-1= 7n + 48
Subtract 48 from both sides
-49= 7n
Divide 7 on both sides so that the only thing remaining on the right sides is the varible n.
Final Answer: -7 = n</span>
Answer:
Total Cost (T)= .80a+1.25c
T= a+c
Step-by-step explanation:
This equation has 2 variables. It also has 2 equations
Answer:
Therefore the area of the quadrilateral =35 cm²
Step-by-step explanation:
Given, the length of one of diagonal of quadrilateral is 10 cm and perpendicular drawn from the opposite vertices to this diagonal are the length of 2.8 cm and 4.2 cm.
A diagonal divided a quadrilateral into two triangle.
Therefore the area of the quadrilateral
= sum of the area of the triangles
cm² [ area of a triangle
]
=35 cm²
