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3 years ago

How many real solutions does a quadratic equation have if its discriminant is negative?

Mathematics

1 answer:

kvasek [131]3 years ago

3 0

Answer:

Step-by-step explanation:

Discriminant is given by

$D = b^{2} -4ac$

If it is negative then in quadratic formula the square root part becomes negative which makes the solution complex.

so there cannot be any real solution if the Discriminant is negative or less than 0 ,

The correct option for the given question is

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If anyone could help answer this question that would be great.

lys-0071 [83]

I believe the answer is 64
First break it off into 2 rectangles on is 4 width and 10 length other is 8 length and 10 width. Then multiple to find area 10•4=40 8•3=24 then add 40+24= 64

3 0

3 years ago

1. What are the intercepts of the equation 2x+3/2y+3z=6

Roman55 [17]

Use
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4 0

3 years ago

Read 2 more answers

Explain how to multiply the following whole numbers 21 x 14

Lesechka [4]

Answer:

$\begin{matrix}\space\space&\textbf{2}&\textbf{1}\\ \times \:&1&\textbf{4}\end{matrix}$

________

$\frac{\begin{matrix}\space\space&\textbf{0}&8&4\\ +&\textbf{2}&1&0\end{matrix}}{\begin{matrix}\space\space&\textbf{2}&9&4\end{matrix}}$

Step-by-step explanation:

Given

$21\:\times \:14$

Line up the numbers

$\begin{matrix}\space\space&2&1\\ \times \:&1&4\end{matrix}$

Multiply the top number by the bottom number one digit at a time starting with the ones digit left(from right to left right)

Multiply the top number by the bolded digit of the bottom number

$\begin{matrix}\space\space&\textbf{2}&\textbf{1}\\ \times \:&1&\textbf{4}\end{matrix}$

Multiply the bold numbers: 1×4=4

$\frac{\begin{matrix}\space\space&2&\textbf{1}\\ \times \:&1&\textbf{4}\end{matrix}}{\begin{matrix}\space\space&\space\space&4\end{matrix}}$

Multiply the bold numbers: 2×4=8

$\frac{\begin{matrix}\space\space&\textbf{2}&1\\ \times \:&1&\textbf{4}\end{matrix}}{\begin{matrix}\space\space&8&4\end{matrix}}$

Multiply the top number by the bolded digit of the bottom number

$\frac{\begin{matrix}\space\space&\textbf{2}&\textbf{1}\\ \times \:&\textbf{1}&4\end{matrix}}{\begin{matrix}\space\space&8&4\end{matrix}}$

Multiply the bold numbers: 1×1=1

$\frac{\begin{matrix}\space\space&\space\space&2&\textbf{1}\\ \space\space&\times \:&\textbf{1}&4\end{matrix}}{\begin{matrix}\space\space&\space\space&8&4\\ \space\space&\space\space&1&\space\space\end{matrix}}$

Multiply the bold numbers: 2×1=2

$\frac{\begin{matrix}\space\space&\space\space&\textbf{2}&1\\ \space\space&\times \:&\textbf{1}&4\end{matrix}}{\begin{matrix}\space\space&\space\space&8&4\\ \space\space&2&1&\space\space\end{matrix}}$

Add the rows to get the answer. For simplicity, fill in trailing zeros.

$\frac{\begin{matrix}\space\space&\space\space&2&1\\ \space\space&\times \:&1&4\end{matrix}}{\begin{matrix}\space\space&0&8&4\\ \space\space&2&1&0\end{matrix}}$

adding portion

$\begin{matrix}\space\space&0&8&4\\ +&2&1&0\end{matrix}$

Add the digits of the right-most column: 4+0=4

$\frac{\begin{matrix}\space\space&0&8&\textbf{4}\\ +&2&1&\textbf{0}\end{matrix}}{\begin{matrix}\space\space&\space\space&\space\space&\textbf{4}\end{matrix}}$

Add the digits of the right-most column: 8+1=9

$\frac{\begin{matrix}\space\space&0&\textbf{8}&4\\ +&2&\textbf{1}&0\end{matrix}}{\begin{matrix}\space\space&\space\space&\textbf{9}&4\end{matrix}}$

Add the digits of the right-most column: 0+2=2

$\frac{\begin{matrix}\space\space&\textbf{0}&8&4\\ +&\textbf{2}&1&0\end{matrix}}{\begin{matrix}\space\space&\textbf{2}&9&4\end{matrix}}$

Therefore,

$\begin{matrix}\space\space&\textbf{2}&\textbf{1}\\ \times \:&1&\textbf{4}\end{matrix}$

________

$\frac{\begin{matrix}\space\space&\textbf{0}&8&4\\ +&\textbf{2}&1&0\end{matrix}}{\begin{matrix}\space\space&\textbf{2}&9&4\end{matrix}}$

6 0

3 years ago

The perimeters of square region S and rectangular region R are equal. If the sides of R are in the ratio 2 : 3, what is the rati

Ksivusya [100]

<h2>Answer:</h2>

The ratio of the area of region R to the area of region S is:

$\dfrac{24}{25}$

<h2>Step-by-step explanation:</h2>

The sides of R are in the ratio : 2:3

Let the length of R be: 2x

and the width of R be: 3x

i.e. The perimeter of R is given by:

$Perimeter\ of\ R=2(2x+3x)$

( Since, the perimeter of a rectangle with length L and breadth or width B is given by:

$Perimeter=2(L+B)$ )

Hence, we get:

$Perimeter\ of\ R=2(5x)$

i.e.

$Perimeter\ of\ R=10x$

Also, let " s " denote the side of the square region.

We know that the perimeter of a square with side " s " is given by:

$\text{Perimeter\ of\ square}=4s$

Now, it is given that:

The perimeters of square region S and rectangular region R are equal.

i.e.

$4s=10x\\\\i.e.\\\\s=\dfrac{10x}{4}\\\\s=\dfrac{5x}{2}$

Now, we know that the area of a square is given by:

$\text{Area\ of\ square}=s^2$

and

$\text{Area\ of\ Rectangle}=L\times B$

Hence, we get:

$\text{Area\ of\ square}=(\dfrac{5x}{2})^2=\dfrac{25x^2}{4}$

and

$\text{Area\ of\ Rectangle}=2x\times 3x$

i.e.

$\text{Area\ of\ Rectangle}=6x^2$

Hence,

Ratio of the area of region R to the area of region S is:

$=\dfrac{6x^2}{\dfrac{25x^2}{4}}\\\\=\dfrac{6x^2\times 4}{25x^2}\\\\=\dfrac{24}{25}$

6 0

3 years ago

Read 2 more answers

The product of two positive integers plus their sum is 95. The integers are relatively prime, and each is less than 20. What is

creativ13 [48]

Let the fist integer be x, the second is x+20
the product of the numbers is:
x(x+20)
the sum of the numbers is:
x+x+20=2x+20
the sum of the above operations will give us:
2x+20+x^2+20x=95
x^2+22x+20=95
this can be written as quadratic to be:
x^2+22x-75=0
solving the above we get:
x=3 and x=-25
but since the integers should be positive, then x=3
the second number is x+20=3+20=23
hence the numbers are:
3 and 23

3 0

3 years ago