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REY [17]
3 years ago
15

What are the possible values of x in 8x2 + 4x = -1?

Mathematics
1 answer:
Westkost [7]3 years ago
4 0

Answer:

The possible values of x are:

x= \frac{-1+\sqrt{3}}{4} \,\,or\,\, x= 0.18

and

x= \frac{-1-\sqrt{3}}{4} \,\, or \,\, x= -0.68

Step-by-step explanation:

8x^2+4x+1=0

Using Quadratic formula to solve this equation:

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\a = 8 \,\, b = 4\,\, c = -1\\Putting \,\, values \,\, in \,\, the\,\, equation\\x= \frac{-4\pm\sqrt{(4)^2-4(8)(-1)}}{2(8)}\\x= \frac{-4\pm\sqrt{16+32}}{16}\\x= \frac{-4\pm\sqrt{48}}{16}\\x= \frac{-4+ \sqrt{48}}{16} \,\, and \,\, x= \frac{-4- \sqrt{48}}{16}\\x= \frac{-1+ \sqrt{3}}{4} \,\, and \,\, x= \frac{-1- \sqrt{3}}{4}

The possible values of x are:

x= \frac{-1+\sqrt{3}}{4} \,\,or\,\, x= 0.18

and

x= \frac{-1-\sqrt{3}}{4} \,\, or \,\, x= -0.68

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Mr. pham cuts a cake into 8 equal slices. then, he cuts every slice in half. how many of the smaller slices does he have. use wo
Katen [24]
8 times 2 equals 16 so the answer would be 16

4 0
3 years ago
Suppose that there are five young women and six young men on an island. Each woman is willing to marry some of the men on the is
pochemuha

Answer:

Based on the current information, the last man remaining unmarried will be a choice between two

Jason marries Anna

Matt marries Elizabeth

Kevin marries Barbara

Larry marries Diane

Carol can marry either Nick or Oscar

Step-by-step explanation:

The basic method to sort which woman will be marrying which man is simple.

We have to extract the choice by comparing choices of each woman side by side and then selecting the least common one and hence sorting one by one.

First, let's name all the women and men

1. Anna

2. Barbara

3. Carol

4. Diane

5. Elizabeth

We will refer to them with the initial letter of their name . A for Anna, B for Barbara, C for Carol, D for Diane, E for Elizabeth.

The men :

1. Jason

2. Larry

3. Matt

4. Kevin

5. Oscar

6. Nick

And they will be referred as J for Jason, L for Larry, M for Matt, K for Kevin, O for Oscar  and N for Nick.

Options for A : J , L ,M

Options for B : K, L

Options for C : J, N , O

Options for D : J, L , N , O

Options for E : J, M

As we can see , no one else wants Kevin other than Barbara so :

Kevin marries Barbara

Moving onwards:

Elizabeth can marry either Jason or Matt,

Meanwhile Anna can marry either Jason , Larry or Matt  

Hence,

Matt marries Eilzabeth

as Anna has two more choices still which are Jason and Larry.

Further, seeing the options for Carol and Diane

Carol can marry either Jason, Nick or Oscar while Diane can marry Larry as well as Jason , Nick and Oscar

We can select Jason as the match for Anna as Larry is also a choice for Diane.

Hence, Jason marries Anna

Now we are left with Larry , Nick and Oscar.

Since Carol can only decide from Nick and Oscar but Diane can decide from Larry, Nick and Oscar. Larry is the ideal option for Diane

so, Larry marries Diane

Now in the end, Carol can either choose between Nick or Oscar.

If she chooses Nick to marry, Oscar is left single

If she chooses Oscar to marry, Nick is left single.

7 0
2 years ago
78×13-8×78+26×78-78=
Amanda [17]

Applying BODMAS (Order of operations rule)

Where: B is for brackets

O is for 'of' which means multiplication

D is for division

M is for multiplication

A is for addition

S is for subtraction

<u>In this case:</u>

We start with addition then we go to addition and finally we subtract.

(78 × 13) - (8 × 78) + (26 × 78) - 78

= 1014 - 624 + 2028 - 78

= 1014 + 2028 - 624 - 78

= 3042 - 624 - 78

=2340

7 0
3 years ago
Read 2 more answers
2=ln(x+7) solve for x
maria [59]

Answer:

e^2 -7 = x

Step-by-step explanation:

2=ln(x+7)

Raise each side to the base of e

e^2 = e^ln (x+7)

The e^ln cancel

e^2 = x+7

Subtract 7 from each side

e^2 -7 = x +7-7

e^2 -7 = x

8 0
3 years ago
What's the answer <br> BchbhbdbhdjbcdjcnjcjndJncjjj
Olenka [21]
Answer: 1,3,6

Explanation:
7 0
3 years ago
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