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RSB [31]
3 years ago
11

Wthat number can be written as 60+3?

Mathematics
2 answers:
Luden [163]3 years ago
8 0
63 can be written as 60+3
igomit [66]3 years ago
7 0
63 can be written as 60+3
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Help again <br> álgebra 1
OLga [1]

Answer:

A (4,6)

ok im going to go through this step by step

4(4)=3(6)-2

16=18-2

16=16 (true solution)

18=3(4)+(6)

18=12+6

18=18 (true solution)

4 0
3 years ago
A Ferris wheel has a radius of 50 feet and its center is 60 feet off the ground how many points on the Ferris wheel are
Vladimir [108]

Answer: 5 feet off the ground

Step-by-step explanation:

3 0
2 years ago
HELPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPP
jolli1 [7]

Answer:

y = 5cos(πx/4) +11

Step-by-step explanation:

The radius is 5 ft, so that will be the multiplier of the trig function.

The car starts at the top of the wheel, so the appropriate trig function is cosine, which is 1 (its maximum value) when its argument is zero.

The period is 8 seconds, so the argument of the cosine function will be 2π(x/8) = πx/4. This changes by 2π when x changes by 8.

The centerline of the wheel is the sum of the minimum and the radius, so is 6+5 = 11 ft. This is the offset of the scaled cosine function.

Putting that all together, you get

... y = 5cos(π/4x) + 11

_____

The answer selections don't seem to consistently identify the argument of the trig function properly. We assume that π/4(x) means (πx/4), where this product is the argument of the trig function.

4 0
3 years ago
Read 2 more answers
Using all or some of the digits 2, 3, 4 and 5, Edward writes
Vesnalui [34]

Using the arrangements formula, considering the given digits, it is found that Edward writes 6 numbers.

<h3>What is the arrangements formula?</h3>

The number of possible arrangements of n elements is given by the factorial of n, that is:

A_n = n!

In this problem, the first digit has to be 5, while the other three are arranged, hence the total number of arrangements is given by:

A_3 = 3! = 6

More can be learned about the arrangements formula at brainly.com/question/25925367

#SPJ1

5 0
1 year ago
Prove the divisibility of the following numbers:
ratelena [41]

Make use of prime factorizations:

16^5+2^{15}=(2^4)^5+2^{15}=2^{20}+2^{15}

Both terms have a common factor of 2^{15}:

16^5+2^{15}=2^{15}\left(2^5+1\right)=2^{15}\cdot33

- - -

The second one is not true! We can write

15^7+5^{13}=(3\cdot5)^7+5^{13}=3^7\cdot5^7+5^{13}

Both terms have a common factor of 5^7:

15^7+5^{13}=5^7\left(3^7+5^6\right)

Since 30=5\cdot6, and 5\mid5^7, we'd still have to show that 5^6(3^7+5^6) is a multiple of 6. This is impossible, because 6=3\cdot2 and there is no multiple of 2 that can be factored out.

4 0
3 years ago
Read 2 more answers
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