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2 years ago

Write the prime factorization of 98

Mathematics

1 answer:

Natasha_Volkova [10]2 years ago

7 0

Answer:

2 * 7^2

Step-by-step explanation:

98 /2 = 49

49 / 7 = 7

2 * 7 * 7 = 98

plz mark brainliest

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How many roots does -2x^2-8=0 have

Aloiza [94]

Answer: two

Step-by-step explanation:

Factor out the equation:

-2( $x^{2}$ + 4) = 0

-2 cannot equal zero, so that means $x^{2}$ - 4 will have to equal to zero in order to get zero as a product

$x^{2}$ - 4 = 0

Factor the equation to get:

(x+2)(x-2) = 0

x = 2, x = -2

The equation has two roots

4 0

3 years ago

PLEASE HELP FAST-----------Passes through the point (-2, 0) and is perpendicular to the line y=-9

miskamm [114]

Answer: x = -2

Step-by-step explanation:

A line perpendicular to y= (anything) is ....... x = ____

Since it must pass through a coordinate x = -2, y = 0

then the equation of the line is: x = -2

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4 years ago

If the sum of $1! + 2! + 3! + \cdots + 49! + 50!$ is divided by $15$, what is the remainder?

kompoz [17]

Answer:

Step-by-step explanation:

All factorials 5 and above are evenly divisible by 15, so have no remainder. Thus, you are interested in ...

mod(1! +2! +3! +4!, 15) = mod(1 +2 +6 +24, 15)

= mod(33, 15) = 3

The remainder is 3.

3 0

4 years ago

Simplify, Rewrite the expression in the form 2n. 2.2.2.2.2.2.2.2 2.2.2.2

Nastasia [14]

Answer: 8/4

Step-by-step explanation: if I’m wrong just watch the video for a hint it helps

6 0

3 years ago

A physics exam consists of 9 multiple-choice questions and 6 open-ended problems in which all work must be shown. If an examinee

katrin [286]

Answer: A) 1260

Step-by-step explanation:

We know that the number of combinations of n things taking r at a time is given by :-

$^nC_r=\dfrac{n!}{(n-r)!r!}$

Given : Total multiple-choice questions = 9

Total open-ended problems=6

If an examine must answer 6 of the multiple-choice questions and 4 of the open-ended problems ,

No. of ways to answer 6 multiple-choice questions

= $^9C_6=\dfrac{9!}{6!(9-6)!}=\dfrac{9\times8\times7\times6!}{6!3!}=84$

No. of ways to answer 4 open-ended problems

= $^6C_4=\dfrac{6!}{4!(6-4)!}=\dfrac{6\times5\times4!}{4!2!}=15$

Then by using the Fundamental principal of counting the number of ways can the questions and problems be chosen = No. of ways to answer 6 multiple-choice questions x No. of ways to answer 4 open-ended problems

= $84\times15=1260$

Hence, the correct answer is option A) 1260

5 0

3 years ago