What is 7.06 divide 0.353

When you solve a word problem involving a quadratic equation which of the following is not a number of solutions can you possibl

Nady [450]

It depends on the value of the discriminant of the quadratic.

If you are familiar with the quadratic formula...

x=(-b±√(b^2-4ac))/(2a)

The part under the radical sign (b^2-4ac) is called the discriminant of the quadratic (of the form ax^2+bx+c)

If the discriminant is less than zero there are no REAL solutions (however there are two imaginary solutions)

If the discriminant is equal to zero, there is just one solution

If the discriminant is greater than zero, there are two real solutions.

So depending on context, if you are to exclude imaginary solutions (which we often do) you can have 0,1, or 2 real solutions.

However you will NEVER have three solutions. (real or imaginary)

6 0

4 years ago

Read 2 more answers

Represente em uma reta numérica os conjuntos indicados abaixo (cada conjunto deve ser representado em uma reta numérica diferent

zmey [24]

La respuesta es C(-2,3) n(3,6)

3 0

3 years ago

–20, - 12,5, 22, . . .<br> An=<br> A50=

marshall27 [118]

Answer:

an = 17n - 46

a50 = 804

Step-by-step explanation:

-29, -12, 5, 22, ...

Subtract each number from the next one.

-12 - (-29) = 17

5 - (-12) = 17

22 - 5 = 17

The common difference is 17. This is an arithmetic sequence which starts with -29, and in which each subsequent value is 17 more than the previous value.

a1 = -29

a2 = -29 + 17

a3 = -29 + 2(17)

a4 = -29 + 3(17)

Notice that for each term, you have -29 and something added to it. What you add to -29 is 17 multiplied by 1 less than the number of the term. For term 1, 1 less than 1 is 0. You add 0 * 17 to -29 and get -29. term 1 is -29. For term 2, 1 less than 2 is 1. You add 1 * 17 to -29 and get -12, etc.

For term n, 1 less than n is n - 1. Add (n - 1) * 17 to -29 to get term n.

an = -29 + 17(n - 1)

This formula can be simplified.

an = -29 + 17n - 17

an = -46 + 17n

an = 17n - 46

a50 = 17(50) - 46

a50 = 804

8 0

3 years ago

Determine whether each of these functions from {a, b, c, d} to itself is one-to-one.

OLga [1]

Answer:

Only function (a) is one-to-one.

Step-by-step explanation:

If every element of the range of the function corresponds to exactly one element of the domain, then the function is called one-to-one.

The elements of a function are

A = {a,b,c,d}

We need to check whether $f\rightarrow A\times A$ is one to one or not.

(a)

f (a) = b, f (b) = a, f (c) = c, f (d) = d

Here, for each value of x, there exist exactly one value of y.

Therefore, this function is one-to-one.

(b)

f (a) = b, f (b) = b, f (c) = d, f (d) = c

Here, the value of function is b for x=a and x=b. For two domains the functions has same range.

Therefore, this function is not one-to-one.

(c)

f (a) = d, f (b) = b, f (c) = c, f (d) = d

Here, the value of function is d for x=a and x=d. For two domains the functions has same range.

Therefore, this function is not one-to-one.

4 0

3 years ago

Find f(g(x)) and g(f(x)) of f(x)=x^2-8, g(x)=2x-5

Zina [86]

$f(x)=x^2-8\\\\g(x)=2x-5\\\\f(g(x))\to\text{instead of x write the equation of the function g(x)}\\\\f(g(x))=(2x-5)^2-8=(2x)^2-2(2x)(5)+5^2-8\\\\=4x^2-20x+25-8=4x^2-20x+17\\\\\text{used}\ (a-b)^2=a^2-2ab+b^2$

$g(f(x))\to\text{instead of x write the equation of the function f(x)}\\\\g(f(x))=2(x^2-8)-5=(2)(x^2)+(2)(-8)-5=2x^2-16-5=2x^2-21$

5 0

3 years ago