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Black_prince [1.1K]
2 years ago
5

Please help will give brainliest

Mathematics
2 answers:
Schach [20]2 years ago
7 0
Your answer is 4s because s+s+s+s
arsen [322]2 years ago
3 0

Answer:

4s

s+s+s+s

Step-by-step explanation:

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Which equation could have been used to create this graph?
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Y=X+9
9 is called y intercept that's where pt starts
The number is front of X is called slope if no number it's 1 so it's 1/1 one up and one right up positive right positive start from 9
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3 years ago
<img src="https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B1000%7D%20%2B%20%5Csqrt%5B3%5D%7B0.027%7D%20-%20%5Csqrt%5B3%5D%7B125%7D" id="
e-lub [12.9K]
Step 1: calculate the cube roots, convert the decimal into a fraction, and calculate the cube roots again

10 + 3 square root 27/1000 - 5

Step 2: calculate the cube root

10 + 3/10 -5

Step 3: subtract the #’s

5 + 3/10

Step 4: calculate and you get

53/10


Answer: 53/10

3 0
3 years ago
Can someone plz solve this for me??
denis-greek [22]

Answer:

-2 x 3 + t = -6

Step-by-step explanation:


8 0
3 years ago
Read 2 more answers
At a café, a cup of coffee costs $2.82. To use exact change, you must hold the number 2.82 in your head while sorting through yo
Stolb23 [73]

Answer:

Working memory

Step-by-step explanation:

Working memory is a system the brain uses  for temporarily storing and managing the information required to carry out complex cognitive tasks. The central executive part of the prefrontal cortex at the front of the brain is responsible for working memory. It serves as a temporary store for short-term memory, where information is kept available while it is needed for current reasoning processes.

So to be successful in holding the number 2.82 in my head while sorting, one must keep the information maintained in short-term storage by using one's working memory.

5 0
3 years ago
Can someone please help me​
astra-53 [7]

Answer:

square 20 has 44 green squares

square 21 has 45 green squares

Step-by-step explanation:

To solve the problem, we need to observe the cases, and determine/define a rule for each case (odd number of sides, or even number of sides).

For square one, we note that the centre square is shared by two diagonals, so we saved one square from the two diagonals.

The side length is 3 for square 1, 4 for square 2, and so on.

Let

n= square number (1, 2,3...)

L = side length (3,4,5...)

G1(n) = function that gives the number of green squares for square n, n=odd

G2(n) = function that gives the number of green squares for square n, n=even

side length, L=n+2   ................(1)

G1(n) = twice the side length less one, as discussed above

G1(n) = 2L-1       now substitute L=n+2

G1(n) = 2(n+2) -1    simplify

G1(n) = 2n + 3

Check:

for n=1, square 1 has 2*1+3 = 5 green squares ... checks

for n=3, square 3 has 2*3+3 = 9... checks

for n=5, square 5 has 2*5+3 = 13 ....checks

For even squares, it is even easier, because

G2(n) = 2L = 2(n+2)

check:

for n=2, square 2 has 2(2+2) = 8 green squares........checks

for n=4, square 4 has 2(4+2) = 12 green squares........checks.

Fincally, we apply our formula to n=20 and n=21

square 20 : G2(20) = 2(n+2) = 2(20+2) = 44 green quars

square 21 : G1(21) = 2n+3 = 2(21)+3 = 45 green squares

5 0
2 years ago
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