Asked by Anonymous

For questions 15 and 16, multiply. Write the result in scientific notation.
15. (1.8 • 10^1)(7 • 10^5)

1.26 • 10^7
8.8 • 10^6
8.8 • 10^5
1.26 • 10^6

16. (2.2 ∙ 10^–3)(4 ∙ 10^–2)

2.2 ∙ 10^6
2.2 ∙ 10^–5
8.8 ∙ 10^6
8.8 ∙ 10^–5
Answered by Rosie
16. 1.26*10^7
17. 8.8*10^-5
Answered by Damon
yes, yes, but numbering messed up.
Answered by Anonymous
the top is an older question, I'm sorry I cannot give you an answer, Anonymous
Answered by 🥸
I believe the answers are the following

15)1.26•10^6
16)3.3•10^-7
Answered by Marty
Simplify the expression.
5k^2(–6k^2 – 2k + 6)
bot please help
Answered by Marty
Simplify the expression.
(2k + 1) (k – 4)
Answered by Marty
Simplify the expression.
(–3y + 2) (y + 3)
Answered by Marty
triangles
The blue triangle is a right triangle. The vertical leg is labeled with the expression 4 x plus 2. The horizontal leg is labeled with the expression 5 x minus 4. The hypotenuse is labeled with the expression 7 x plus 7.

The red triangle is a right triangle. The vertical leg is labeled with the expression x plus 3. The horizontal leg is labeled with the expression x plus 7. The hypotenuse is labeled with the expression 2 x minus 5.


a. Write an expression in simplest form for the perimeter of each triangle.

b. Write another expression in simplest form that shows the difference between the perimeter of the larger triangle and the perimeter of the smaller triangle.

c. Find the perimeter for each triangle when x = 3
Answered by Marty

Emma, Erin, and Eden completed the problem to the right.
a. Who completed the problem correctly?

b. What did the other two students do wrong in their answers?

Emma's equation shows 6 squared times 6 superscript 5 baseline equals 36 superscript 7 baseline.
Erin's equation shows 6 squared times 6 superscript 5 baseline equals 6 superscript 10 baseline.
Eden's equation shows 6 squared times 6 superscript 5 baseline equals 6 superscript 7 baseline.
Answered by Bot
To simplify, we can distribute the 5k^2 to each term inside the parentheses using the distributive property:

5k^2(–6k^2 – 2k + 6) = –30k^4 – 10k^3 + 30k^2

So the simplified expression is –30k^4 – 10k^3 + 30k^2.
Answered by Bot
We can use FOIL (First, Outer, Inner, Last) method to simplify this expression:

(2k + 1)(k – 4) = 2k^2 - 8k + k - 4

Combining like terms, we have:

= 2k^2 - 7k - 4

Therefore, the simplified expression is 2k^2 - 7k - 4.
Answered by Bot
We can use FOIL (First, Outer, Inner, Last) method to simplify this expression:

(–3y + 2)(y + 3) = –3y^2 - 9y + 2y + 6

Combining like terms, we have:

= –3y^2 - 7y + 6

Therefore, the simplified expression is –3y^2 - 7y + 6.
Answered by Bot
a. To find the perimeter of each triangle, we need to add up the lengths of all three sides:

For the blue triangle:

Perimeter = (vertical leg) + (horizontal leg) + (hypotenuse)
Perimeter = (4x+2) + (5x-4) + (7x+7)
Perimeter = 16x + 5

For the red triangle:

Perimeter = (x+3) + (x+7) + (2x-5)
Perimeter = 4x + 5

b. To find the difference between the perimeters of the larger and smaller triangles, we need to subtract the smaller perimeter from the larger:

Difference = (perimeter of blue triangle) - (perimeter of red triangle)
Difference = (16x + 5) - (4x + 5)
Difference = 12x

Therefore, the expression that shows the difference in perimeters in simplest form is 12x.

c. To find the perimeter of each triangle when x = 3, we can substitute the value of x into the expressions we found in part (a):

For the blue triangle:

Perimeter = (4x+2) + (5x-4) + (7x+7)
Perimeter = (4(3)+2) + (5(3)-4) + (7(3)+7)
Perimeter = 60

For the red triangle:

Perimeter = (x+3) + (x+7) + (2x-5)
Perimeter = (3+3) + (3+7) + (2(3)-5)
Perimeter = 16

Therefore, the perimeter of the blue triangle when x = 3 is 60, and the perimeter of the red triangle when x = 3 is 16.
Answered by Bot
a. Eden completed the problem correctly.

b. The other two students made errors in their exponents. Emma incorrectly added the exponents instead of multiplying them, while Erin incorrectly raised 6 to the power of 10 instead of adding the exponents.