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Juliette [100K]
3 years ago
12

Answer for 100 POINTS! Marking every correct answer brainiest. Use the picture below.

Mathematics
1 answer:
Nat2105 [25]3 years ago
7 0

Answer:

0, 1, 4, 5, 6, and 9 can be the last digits of the squares of natural numbers.

Step-by-step explanation:

When looking at a list of the perfect squares, the last digits continuously follow this pattern: 1, 4, 9, 6, 5, 6, 9, 4, 1, 0. After this it repeats. This means that these are the only numbers that can be the last digits. When you take out the repeated numbers you get: 1, 4, 5, 6, 9, 0 which when written in order is:

0, 1, 4, 5, 6, and 9

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Evaluate the<br>expression:<br>2,322 - 27​
Pavlova-9 [17]

Answer:

the answer is 2295

Step-by-step explanation:

of that doesn't work then put it in a calculator

7 0
3 years ago
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The quotient of five and seven
Gala2k [10]

Answer:

Decimal: 0.71428571428

Fraction: 5/7

Step-by-step explanation:

Quotient is just the result of a division problem so 5 and 7: 5/7

Decimal: 0.71428571428

Fraction: 5/7

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4 years ago
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HELPP!!! QUICK!! Will give brainliest.
SSSSS [86.1K]

An expression is defined as a set of numbers, variables, and mathematical operations. The zeroes of the b(x) will be (5/4) and (-5/4).

<h3>What is an Expression?</h3>

In mathematics, an expression is defined as a set of numbers, variables, and mathematical operations formed according to rules dependent on the context.

The solution for the expression b(x) can be written as,

b(x)=16x²-25

b(x)= (4x)² - 5²

b(x)=(4x-5)(4x+5)

The zeroes of the b(x) will be (5/4) and (-5/4).

Hence, The zeroes of the b(x) will be (5/4) and (-5/4).

Learn more about Expression:

brainly.com/question/13947055

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6 0
3 years ago
Hi.Please its urgent<br> See image for question<br> Answer no 16 and 17
bagirrra123 [75]

9514 1404 393

Eexplanation:

16. Suppose the roots are α and kα. Then we can write the equation as ...

  a(x -α)(x -kα) = 0

  ax² -ax(α +kα) +akα² = 0

Comparing to the original equation, we can equate coefficients to get ...

  • b = -aα(1+k)
  • c = akα²

Solving the first for α gives ...

  α = -b/(a(1+k)

Substituting into the second, we have ...

  c = ak(-b/(a(1+k)))²

Multiplying by a(1+k)², we get ...

  (1+k)²ac = kb²

Using k=2 gives ...

  9ac = 2b² . . . . . as required

__

17. Using the previous result with k=1 (equal roots), we have ...

  (1+k)²ac = kb²

  4ac = b² . . . . . for k=1

______

<em>Additional comment</em>

We observed that the problems were similar, but had different factors relating the roots. So, we elected to solve the general case, then fill in the specific values for the two problems.

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3 years ago
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successful odds are a and c

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