Solve the equation by completing the square. round to the nearest hundredth if necessary. x^2+3x-5=0

The cost of 8 exercise book is rs 400.

grandymaker [24]

Answer:

e

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

A bag contains red marbles, white marbles, and blue marbles. Randomly choose two marbles, one at a time, and without replacement

dsp73

Answer:

$P(First\ White\ and\ Second\ Blue) = \frac{3}{28}$

$P(Same) = \frac{67}{210}$

Step-by-step explanation:

Given (Omitted from the question)

$Red = 7$

$White = 9$

$Blue = 5$

Solving (a): $P(First\ White\ and\ Second\ Blue)$

This is calculated using:

$P(First\ White\ and\ Second\ Blue) = P(White) * P(Blue)$

$P(First\ White\ and\ Second\ Blue) = \frac{n(White)}{Total} * \frac{n(Blue)}{Total - 1}$

<em>We used Total - 1 because it is a probability without replacement</em>

So, we have:

$P(First\ White\ and\ Second\ Blue) = \frac{9}{21} * \frac{5}{21 - 1}$

$P(First\ White\ and\ Second\ Blue) = \frac{9}{21} * \frac{5}{20}$

$P(First\ White\ and\ Second\ Blue) = \frac{9*5}{21*20}$

$P(First\ White\ and\ Second\ Blue) = \frac{45}{420}$

$P(First\ White\ and\ Second\ Blue) = \frac{3}{28}$

Solving (b) $P(Same)$

This is calculated as:

$P(Same) = P(First\ Blue\ and Second\ Blue)\$ $or\ P(First\ Red\ and Second\ Red)\ or\ P(First\ White\ and Second\ White)$

$P(Same) = (\frac{n(Blue)}{Total} * \frac{n(Blue)-1}{Total-1})+(\frac{n(Red)}{Total} * \frac{n(Red)-1}{Total-1})+(\frac{n(White)}{Total} * \frac{n(White)-1}{Total-1})$

$P(Same) = (\frac{5}{21} * \frac{4}{20})+(\frac{7}{21} * \frac{6}{20})+(\frac{9}{21} * \frac{8}{20})$

$P(Same) = \frac{20}{420}+\frac{42}{420} +\frac{72}{420}$

$P(Same) = \frac{20+42+72}{420}$

$P(Same) = \frac{134}{420}$

$P(Same) = \frac{67}{210}$

6 0

2 years ago

Solve for the variable <img src="https://tex.z-dn.net/?f=v_i" id="TexFormula1" title="v_i" alt="v_i" align="absmiddle" class="la

Mademuasel [1]

Answer:

±sqrt( (vf) ^2 -2ad)) = vi

Step-by-step explanation:

(vf) ^2 = (vi)^2 +2ad

Subtract 2ad from each side

(vf) ^2 -2ad = (vi)^2 +2ad -2ad

(vf) ^2 -2ad = (vi)^2

Take the square root of each side

±sqrt((vf) ^2 -2ad)) = sqrt((vi)^2)

±sqrt( (vf) ^2 -2ad)) = vi

3 0

3 years ago

Determine if the relation represents a function. (2,1) (4,-7) (6,0) (4,-1) (2,9)

Nimfa-mama [501]

Answer:

It isn't a function

Step-by-step explanation:

2 and 4 both have 2 x values, and on a graph, a function cannot have more than 1 x value.

4 0

2 years ago

Solve the inequality 2x - 3 < x + 2 ≤ 3x + 5. Show your work.

grandymaker [24]

Answer:

-3/2 ≤ x < 5

Step-by-step explanation:

Begin the solution of this inequality 2x - 3 < x + 2 ≤ 3x + 5 by combining the constants: Subtracting 2 from 2x - 3, x + 2 and 3x + 5 yields:

2x - 5 < x ≤ 3x + 3.

Focus on the first part of this inequality, 2x - 5 < x simplifies to x - 5 < 0 if we subtract x from both sides. Then x < 5 is part of the solution.

Focus now on the second part of this inequality: x ≤ 3x + 3. Subtracting x from both sides yields 0 ≤ 2x + 3, or -3 ≤ 2x, or 2x ≥ -3, or x ≥ -3/2.

Re-combining these two separate inequalities results in:

-3/2 ≤ x < 5 (equivalent to -1.5 ≤ x < 5).

Example: suppose we choose a test number from this interval. Does this test number satisfy the original equation? Choose x = 1.

Is 2(1) - 3 < 1 + 2 ≤ 3(1) + 5 true?

Is 2 - 3 < 3 ≤ 8 true? YES. So x = 1 is part of the solution set -3/2 ≤ x < 5.

5 0

2 years ago