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Mazyrski [523]
3 years ago
8

Context does not dictate when to overestimate and when to underestimate always round down true or false

Mathematics
2 answers:
Sphinxa [80]3 years ago
8 0

Answer: A, Speaking of context, it would have been nice to get some.


Step-by-step explanation:


konstantin123 [22]3 years ago
4 0

Answer:

False

Step-by-step explanation:

If a situation calls for something to be "at least" a number we would typically round up.  For example, if we have a problem that states we need to buy at least $50 worth of items, and solving the problem we get an answer of

x ≥ 6.8,

we would need to round up to 7 items.  6 items is less than 6.8, so it would not work.  However, 7 items is greater than 6.8, so it would work.

If a situation calls for something to be "at most" a number, we would typically round down.  For example, if we had a problem where we could buy at most $100 worth of items, and solving it we get

x ≤ 4.8,

we would round to 4.  We would not have enough money to buy 5 items, so we cannot round up.  However, we can buy 4 items, so we round down to 4.

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Arden has 15 cards and 3 of those cards are red. What percent of her cards are red? *​
Kisachek [45]

Answer: 20%

Step-by-step explanation: 3/15 = 1/5 = .20 = 20%

5 0
4 years ago
the recipe calls for 2 3/4 cups of tomatoes how many cups of tomatoes written as a fraction greater than one are used in the rec
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11/4 cups of tomatoes.
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Write -9t+9 as the product of two factors in a way that is not shown in the Example. Explain how you found it.
adoni [48]

Answer:

9·(-t - 1)

Step-by-step explanation:

9·(-t - 1) is this expression in factored form (product of two factors)

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3 years ago
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placing 12 tiles side by side, Pete can make a 12-by-1 rectangle. Name two other rectangles Pete can make with 12 tiles
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6x2
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4 years ago
Let f be a functions of degree 4 whose coefficients are real numbers: two of its zeros are - 3 and 4 - i. Explain why one of the
kvasek [131]

Answer:

Step-by-step explanation:

We have the following theorem, if f is a polynomial with real coefficients, we can factor it completely in factors of the degree at most 2.

Consider first a polynomial of degree two, hence it is a polynomial of the form ax^2+bx+c. The cuadratic formula tells us that the solutions are of the form

x = \frac{-b\pm \sqrt[]{b^2-4ac}}{2a}.

Note that square root, over the reals, tells us that the are only real solutions if b^2-4ac \geq 0. If that is not the case, say it's negative, the solution are complex. Then, the solutions are of the form

x = \frac{-b \pm i \sqrt[]{4ac-b^2}}{2a}. NOte that this means that if we have a complex number of the form a+bi that is a solution, then the number a-bi (who is called the complex conjugate) is also a solution.

Recall that when we have a polynomial f(x) whose a zero is the number c, then we can factor f as follows f(x) = (x-c) * p(x) where p(x) is another polynomial of lesser degree .

So far, we know that -3 and 4-i are zeros of the function f. Note that we are missing two zeros. But, since complex numbers are zeros of polynomial only by pairs (that is the number and its conjugate are zeros), then, we must have that one of the missing 2 zeros is a real number. We have 4-i as a zero, then, its complex conjugate must be also a zero, i.e 4+i is a zero.

8 0
4 years ago
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