Answer:
the area is 20 in
Step-by-step explanation:
3 times 4 is 12. since we used 3 inches out of the 11, we can only multiply the 8 inches with 1. 8 times 1 is 8. 12+8=20.
The weight of the kitten will be no less than the weight of the puppy in 1.4 + 1.2x ≥ 2.8 + 0.5x week.
<h3>What is Inequality?</h3>
Statement of an order relationship—greater than, greater than or equal to, less than, or less than or equal to—between two numbers or algebraic expressions.
Given:
A puppy weighs 2.8 pounds at birth and gained about 0.5 pounds per week.
A kitten weighs 1.4 pounds at birth and gains about 1.2 pounds per week.
puppy: 2.8 + 0.5x
Kitten: 1.4 + 1.2x
As, kitten is no less than puppy.
Kitten ≥ puppy
1.4 + 1.2x ≥ 2.8 + 0.5x
hence, the weight of the kitten will be no less than the weight of the puppy in 1.4 + 1.2x ≥ 2.8 + 0.5x week.
Learn more about inequality here:
brainly.com/question/23575974
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SOLUTION:
GCF = 8
56 + 24x = 8 ( 7 + 3x )
Hope this helps! :)
Answer:
45.
Step-by-step explanation:
To solve this equation, we must use PEMDAS...
The operations listed first must be solved first before we can continue.
Parentheses
Exponents
Multiplication
Division
Addition
Subtraction
Using this, we find that we must complete the multiplication and division before we add or subtract.
20 / 2 - 10 + 3 * 15
= (20 / 2) - 10 + (3 * 15)
= 10 - 10 + 45
= 0 + 45
= 45
Hope this helps!
Answer:
f(x, y) = Sin(x*y)
We want the second order taylor expansion around x = 0, y = 0.
This will be:

So let's find all the terms:
Remember that:


f(0,0) = sin(0*0) = 1.





Then we have that the taylor expansion of second order around x = 0 and y = 0 is:
sin(x,y) = x*y + x*y + x*y = 3*x*y