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3 years ago

For what real numbers x is the value of x2 – 6x + 9 negative

2 answers:

7 0

This is the concept of quadratic equations. The equation has the original form of y=ax^2+bx+c. Our equation is given by y=x^2-6x+9. Since our leading coefficient is positive, then it means that our function will not be negative at any point. In other words, the range is [0,∞). Hence we conclude that the function will be no where negative.

Nutka1998 [239]3 years ago

3 0

A real numbers are numbers that can be found on a number line, they include both rational and irrational numbers.
We can first solve the equation;
x² - 6x + 9 =0, using completing square method;
x²- 6x + (-3)² = -9 + (-3)²
(x-3)² = 0
x = 3
therefore, both values of x are 3, thus the expression has no real number x is negative.

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3 years ago

Convert the equation r = 12 sin theta to Cartesian coordinates. Describe the resulting curve.

VLD [36.1K]

Answer:

Step-by-step explanation:

Given that there is a polar equation as

$r=12 sin \theta$

This has to be converted into cartesian.

We know the conversion is

$r^2 =x^2+y^2 \\tan \theta = \frac{y}{x}$

Using this we can say that

$sin^2 \theta = 1-cos^2 \theta \\= 1-\frac{1}{sec^2 \theta} \\=1-\frac{1}{1+tan^2 \theta} \\=1-\frac{1}{1+\frac{y^2}{x^2} } \\=1-\frac{x^2}{x^2+y^2} \\=\frac{y^2}{y^2+y^2}$

$\sqrt{x^2+y^2} =12(\frac{y}{\sqrt{x^2+y^2} } \\x^2+y^2 =12y\\x^2+y^2-12y+36 = 36\\x^2+(y-6)^2 = 6^2$ }

Circle with centre (0,6) and radius 6.

7 0

3 years ago

Please help, super hard question for me!

dem82 [27]

Answer:

Step-by-step explanation:

When using the substitution method we use the fact that if two expressions y and x are of equal value x=y, then x may replace y or vice versa in another expression without changing the value of the expression.

Solve the systems of equations using the substitution method

{y=2x+4

{y=3x+2

We substitute the y in the top equation with the expression for the second equation:

2x+4 = 3x+2

4−2 = 3x−2

2=== = x

To determine the y-value, we may proceed by inserting our x-value in any of the equations. We select the first equation:

y= 2x + 4

We plug in x=2 and get

y= 2⋅2+4 = 8

The elimination method requires us to add or subtract the equations in order to eliminate either x or y, often one may not proceed with the addition directly without first multiplying either the first or second equation by some value.

Example:

2x−2y = 8

x+y = 1

We now wish to add the two equations but it will not result in either x or y being eliminated. Therefore we must multiply the second equation by 2 on both sides and get:

2x−2y = 8

2x+2y = 2

Now we attempt to add our system of equations. We commence with the x-terms on the left, and the y-terms thereafter and finally with the numbers on the right side:

(2x+2x) + (−2y+2y) = 8+2

The y-terms have now been eliminated and we now have an equation with only one variable:

4x = 10

x= 10/4 =2.5

Thereafter, in order to determine the y-value we insert x=2.5 in one of the equations. We select the first:

2⋅2.5−2y = 8

5−8 = 2y

−3 =2y

−3/2 =y

y =-1.5

6 0

2 years ago

A bakery packages 868 cupcakes into 31 boxes. The same number of cupcakes are pput into each box. How many cupcakes are in each

ratelena [41]

To find this you would divide 868cupcakes by 31boxes witches is 28cupcakes in each box
hope this helps good luck have a good nite hope u enjoy ur the rest of ur weekend

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4 years ago

Look at this graph:

Lorico [155]

Y = 30/20x + 20
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3 years ago

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