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3 years ago

What is the quotient of (-8x^6/4x^-3)? A.)x^9/32 B.)4x^9 C.)-12/x^2 D.)–2x^9

Mathematics

1 answer:

Korolek [52]3 years ago

4 0

Answer: D

Explanation:

$\frac{-8x^{6}}{4x^{-3}}$

= $\frac{-8}{4} x^{6-(-3)}$

= -2x⁹

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(Picture) MULTIPLYING POLYNOMIALS AND SIMPLIFYING EXPRESSIONS PLEASE HELP!!!!

Mrrafil [7]

Answer:

-32x² +8x -8
See below about the process

Step-by-step explanation:

The simplified expression is ...

3x(x -12x) +3x² -2(x -2)²

= 3x(-11x) +3x² -2(x² -4x +4) . . . . . simplify contents of parentheses

= -33x² +3x² -2x² +8x -8 . . . . . . . .eliminate parentheses

= (-33 +3 -2)x² +8x -8 . . . . . . . . . . group the coefficients of like terms

= -32x² +8x -8

8 0

3 years ago

Read 2 more answers

A: What is the smallest increment on the voltmeter

oee [108]

Answer:

RESOLUTION

Step-by-step explanation:

Resolution is the smallest increment a tool can detect and display. For a nonelectrical example, consider two rulers.

4 0

3 years ago

assume that when adults with smartphones are randomly selected 15 use them in meetings or classes if 15 adult smartphones are ra

Tanzania [10]

Answer:

The probability that at least 4 of them use their smartphones is 0.1773.

Step-by-step explanation:

We are given that when adults with smartphones are randomly selected 15% use them in meetings or classes.

Also, 15 adult smartphones are randomly selected.

Let X = Number of adults who use their smartphones

The above situation can be represented through the binomial distribution;

$P(X = r) = \binom{n}{r}\times p^{r} \times (1-p)^{n-r} ; n = 0,1,2,3,.......$

where, n = number of trials (samples) taken = 15 adult smartphones

r = number of success = at least 4

p = probability of success which in our question is the % of adults

who use them in meetings or classes, i.e. 15%.

So, X ~ Binom(n = 15, p = 0.15)

Now, the probability that at least 4 of them use their smartphones is given by = P(X $\geq$ 4)

P(X $\geq$ 4) = 1 - P(X = 0) - P(X = 1) - P(X = 2) - P(X = 3)

= $1- \binom{15}{0}\times 0.15^{0} \times (1-0.15)^{15-0}-\binom{15}{1}\times 0.15^{1} \times (1-0.15)^{15-1}-\binom{15}{2}\times 0.15^{2} \times (1-0.15)^{15-2}-\binom{15}{3}\times 0.15^{3} \times (1-0.15)^{15-3}$

= $1- (1\times 1\times 0.85^{15})-(15\times 0.15^{1} \times 0.85^{14})-(105 \times 0.15^{2} \times 0.85^{13})-(455 \times 0.15^{3} \times 0.85^{12})$

= 0.1773

3 0

3 years ago

How did the Germanic tribes govern thier kingdoms

Verdich [7]

The Germanic tribes were governed by unwritten costumes.

3 0

3 years ago

Triangles ΔABC ≅ ΔBAD so that C and D lie in the opposite semi-planes of segment AB. Prove that segment CD bisects segment AD.

kolbaska11 [484]

Answer:

See explanation

Step-by-step explanation:

Triangles ΔABC and ΔBAD are congruent. So,

AB ≅ BA;
AC ≅ BD;
BC ≅ AD;
∠ABC ≅ ∠BAD;
∠BCA ≅ ∠ADB;
∠CAB ≅ ∠DBA.

Consider triangles AEC and BED. In these triangles,

AC ≅ BD;
∠EAC ≅ ∠EBD (because ∠CBA ≅ ∠BAD);
∠AEC ≅ ∠BED (as vertical angles).

So, ΔAEC ≅ ΔBED. Thus,

AE ≅ EB.

This means that segment CD bisects segment AD.

6 0

3 years ago