Find each product.
{ this means the square root sign.
1. (3+2{20)(5+3{45)
2. The length and width of the rectangle are x-5 and 2x-6
a. Write a trinomial that represents that area of the rectangle.
b. What is the smallest integer value of x that will give positive values for the length and the width of the rectangle?
This would really be
(5SQRT20)(8SQRT45)
ROunded
(22.4) (53.7)
IM not sure if you have to mulitply these, but if you do.....this would be youe answers
1202.88
Hope this helps.
I am talking about Radicals, not actaully calculating the square roots.
My book says that the answer is
195 + 47*{5
I just don't get how they got that answer.
(3+2√20)(5+3√45)
multiply it using FOIL
= 15 + 9√45 + 10√20 + 6√900
= 15 + 27√5 + 20√5 + 180
= 195 + 47√5
for the second, just multiply the two expression, and again use FOIL
(x-5)(2x-6)
= .....
For the third part, look at x-5
isn't that positive for all values of x > 5 ???
what about the 2x-6 ??
spce./ the diameter of mars is about 9.4 kilometers. write 9.4 as a product. then find the value of the product.
To find each product, we can follow these steps:
1. (3+2{20)(5+3{45)
First, let's simplify the expressions inside the square root:
√20 = 2√5
√45 = 3√5
Now, we can substitute these simplified expressions back into the original equation:
(3+2√5)(5+3√5)
To find the product, we can use the distributive property:
(3)(5) + (3)(3√5) + (2√5)(5) + (2√5)(3√5)
Simplifying further:
15 + 9√5 + 10√5 + 6(√5)^2
Since (√5)^2 is equal to 5:
15 + 9√5 + 10√5 + 6(5)
= 15 + 19√5 + 30
= 45 + 19√5
2. The length and width of the rectangle are x-5 and 2x-6.
a. To find the area of the rectangle, we can simply multiply the length by the width:
Area = (x-5)(2x-6)
Applying the distributive property:
Area = x(2x) - x(6) - 5(2x) + 5(6)
Simplifying further:
Area = 2x^2 - 6x - 10x + 30
Combining like terms:
Area = 2x^2 - 16x + 30
Therefore, the trinomial that represents the area of the rectangle is 2x^2 - 16x + 30.
b. To find the smallest integer value of x that will give positive values for the length and width of the rectangle, we need to set the expressions for length and width greater than zero and solve for x.
Setting x-5 > 0 and 2x-6 > 0:
x > 5 and x > 3
Since we want the smallest integer value that satisfies both conditions, we choose the larger of the two, which is x > 5.
Therefore, the smallest integer value of x that will give positive values for the length and width of the rectangle is x = 6.