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3 years ago

Which shows the expression below simplified? 0.00048 ÷ (6 × 10-4)

1 answer:

5 0

0.00048 ÷ (6 × 10-4) should be rewritten as 0.00048 ÷ (6 × 10^(-4) ). The symbol " ^ " denotes exponentiation. Without parentheses here, it appears that you want to subtract 4 from 10, which is inappropriate.

First focus on the " (6 × 10^(-4) )" This is equivalent to 6 / (10^4). To divide 0.00048 by 6 / (10^4), invert the divisor and then multiply:

0.00048 * (10^4 / 6) = 0.00008*10^4, or 8*10^(-5)*10^(4) = 8*10^(-1) = 8/10 = 0.8.

Note that all of the possible answer choices need to be re-written. 10-1, for example, must be re-written as 10^(-1).

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Suppose y varies inversely with x, and y = 25 when x = 1/5. What is the value of y when x = 5 ?

34kurt

Answer:

y = 1

Step-by-step explanation:

Given that y varies inversely with x then the equation relating them is

y = $\frac{k}{x}$ ← k is the constant of variation

To find k use the condition y = 25 when x = $\frac{1}{5}$ , then

25 = $\frac{k}{\frac{1}{5} }$ = 5k ( divide both sides by 5 )

5 = k

y = $\frac{5}{x}$ ← equation of variation

When x = 5, then

y = $\frac{5}{5}$ = 1

8 0

3 years ago

Find K such that the points (-2, 5), (2, 8) and (6, K) are collinear.

damaskus [11]

Answer:

k = 11

Step-by-step explanation:

let the points be A(-2,5), B(2,8), C(6,K).

for the points to be collinear slope of line AB maust be equal to line BC.

slope of line AB = $\frac{8-5}{2+2}=\frac{3}{4}$

slope of line BC = $\frac{k-8}{6-2}=\frac{k-8}{4}$

therefore $\frac{k-8}{4}= \frac{3}{4}$

therefore k = 8 + 3

k = 11

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3 years ago

Help with Pythagorean theorem

OLEGan [10]

Answer:

Step-by-step explanation:

(27*27)+(36*36) = 2025

square root 2025 and you get 45

to check 45*45=2025

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3 years ago

Find the axis of symmetry for y=3x^2 +5x-7

slamgirl [31]

A = the number before x^2 = 3
b = the number before x = 5
To find the axis of symmetry: -b/2a
x = -5/6

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3 years ago

Let R(ABCDE) be a relation in Boyce-Codd Normal Form (BCNF). If AC is the only key for R, identify each of these FD's from the f

Goshia [24]

Given that R(ABCDE) is in Boyce-Codd normal form.

And AB is the only key for R.

Definition

A relational nontrivial Schema R is in BCNF if FD (X-A) holds in R, Super key of R. whenever then X is

Given that AB is the only key for R.

ABC E (Yes).

check if ABC is a Super key. AB is a key, ABC is A B C E is in BONE a super key.

2) ACE B

(NO). no Check if ACE As there is ACE is not a Super key? AB in Super key. ACE.

ACE B

Boyce-Codd Normal Form not in BENE (NO)

3) ACDE → B (NO)

check if is a super key. ACDE

As ACDE there is not any AB Tn ACDE. a super key.

ACDEB is not in BCNF.

4) BS → C → (NO)

As there is no AB in BC ~. B(→ not in BCNF

BC is not a super key.

5) ABDE (Yes).

Since AB is a key.

ABO TS a super key.

.. ABDE → E is in BCNF

Let R(ABCDE) be a relation in Boyce-Codd Normal Form (BCNF). If AB is the only key for R, identify each of these FDs from the following list. Answer Yes or No and explain your answer to receive points.

1. ABC E

2. ACE B

3. ACDE B

4. BC C

5. ABD E

Learn more about Boyce-Codd Normal Form at

brainly.com/question/14299832

#SPJ4

6 0

1 year ago