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3 years ago

What is the definition of 'distributive property'?

Mathematics

2 answers:

Mademuasel [1]3 years ago

8 0

This is what my math teacher had me write down!

Crazy boy [7]3 years ago

8 0

answer : distribute property lets you multiply a sum by multiplying each addend separately and then add the products.

example : 5(6+2)= 5×6 +5×2

= 30. + 10

=40

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Help help help please homework is killing me right now

Veronika [31]

The domain of a function is the set of input or argument values for which the function is real and defined.

So, for the given function to be defined, we need to find the possible values for which the values of x makes the square root to be positive.

That is;

-9 -5x ≥ 0

Now, let's solve for x

Add 9 to both-side of the equation

-5x ≥ 9

Divide both-side by -5

x ≤ -9/5

Therefore, the domain of the function can be represented in interval notation as: ( - ∞ , -9/5]

8 0

2 years ago

The vertex of the parabola below is at the point (1, 5), and the point (2, 8) is on the parabola. What is the equation of the pa

N76 [4]

Answer: B

Step-by-step explanation:

the parabola having (1,5) as vertex have as equation:

y=k*(x-1)²+5

It is passing through (2,8)

8=k*(2-1)²+5 ==> k+5=8 ==> k=3

equation is y=3(x-1)²+5

Answer B

6 0

3 years ago

Find the equation, (f(x) = a(x - h)2 + k), for a parabola containing point (2, -1) and having (4, -3) as a vertex. What is the s

Nataliya [291]

Answer:

$f(x)=\frac{1}{2}x^2-4x+5$

Step-by-step explanation:

A parabola is written in the form

$f(x)=a((x-h)^2+k)$ (1)

where:

$h$ is the x-coordinate of the vertex of the parabola

$ak$ is the y-coordinate of the vertex of the parabola

$a$ is a scale factor

For the parabola in the problem, we know that the vertex has coordinates (4,-3), so we have:

$h=4$ (2)

$ak=-3$

From this last equation, we get that $a=\frac{-3}{k}$ (3)

Substituting (2) and (3) into (1) we get the new expression:

$f(x)=-\frac{3}{k}((x-4)^2+k) = -\frac{3}{k}(x-4)^2 -3$ (4)

We also know that the parabola contains the point (2,-1), so we can substitute

x = 2

f(x) = -1

Into eq.(4) and find the value of k:

$-1=-\frac{3}{k}(2-4)^2-3\\-1=-\frac{3}{k}\cdot 4 -3\\2=-\frac{12}{k}\\k=-\frac{12}{2}=-6$

So we also get:

$a=-\frac{3}{k}=-\frac{3}{-6}=\frac{1}{2}$

So the equation of the parabola is:

$f(x)=\frac{1}{2}((x-4)^2 -6)$ (5)

Now we want to rewrite it in the standard form, i.e. in the form

$f(x)=ax^2+bx+c$

To do that, we simply rewrite (5) expliciting the various terms, we find:

$f(x)=\frac{1}{2}((x^2-8x+16)-6)=\frac{1}{2}(x^2-8x+10)=\frac{1}{2}x^2-4x+5$

6 0

3 years ago

Which expression is equivalent to cos(3x) sin x?

Drupady [299]

The answer is A)1/2[sin(4x)+sin(2x)]

4 0

3 years ago

Find the measure of the hands of a clock at 12:38

Vinvika [58]

I would say 12 hours and 38 min

8 0

3 years ago