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3 years ago

Find the nonpermissible replacement for a in this expression 6/-5a

1 answer:

4 0

Find the nonpermissible replacement for a in this expression 6/-5a

Denominator cannot be zero or else undefined so set the denominator to zero and solve for the a that will cause the denominator to be zero.
-5a=0
-5/-5x=0/-5
x=0

Nonpermissable replacement is 0.

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erma4kov [3.2K]

$72.

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3 years ago

The numbers 1 through 50 are written with red marker on slips of paper while the numbers 51 through 100 are written with blue ma

allsm [11]

The correct answer for the question that is being presented above is this one:

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2 years ago

The equation 9 x 2 − 12 x + 4 = 0 9 x 2 − 12 x + 4 = 0 has how many different solutions?

Hatshy [7]

We have the equation:

$9x^2-12x+4=0$

so:

$a=9\qquad b=-12\qquad c=4$

and:

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2 years ago

Hannah is 2 times her sister’s age. The sum of their ages is no more than 18 years. 1) Write an inequality that can be used to r

svetoff [14.1K]

Answer:

1) $x+2x\leq 18$ .

2) $6$ years

Step-by-step explanation:

Let Hanna's sister be $x$ years old.

Then Hanna will be $2x$ years old.

The sum of their ages will be $x+2x$ years.

The statement "no more than 18" means the sum of their ages is less than or equal to 18.

The required inequality is;

$x+2x\leq 18$ .

The inequality we just found is

$x+2x\leq18$ .

We simplify to get;

$3x\leq 18$ .

we divide both sides by $3$ to get;

$x\leq 6$ .

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3 years ago

Twelve basketball players, whose uniforms are numbered 1 through 12, stand around the center ring on the court in an arbitrary a

Ksju [112]

Answer:

Shown

Step-by-step explanation:

Given that twelve basketball players, whose uniforms are numbered 1 through 12, stand around the center ring on the court in an arbitrary arrangement.

Let us consider consecutive numbers in this set.

$1+2+3 =6$

After this we find the totals are more than 20.

When 1 to 12 are arbitrarily arranged, there are chances that numbers from 6 and above are having consecutive numbers.

These totals are greater than 20

Hence shown that some three consecutive players have the sum of their numbers at least 20.

(i.e. starting from if we take)

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3 years ago