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3 years ago

10.3 = 0.6y How do I find the answer to this on please

1 answer:

3 0

To solve for y, you want to have all unknowns on one side and knowns on the other. to do this, we can divide both sides by 0.6 so that the knows are on the second side and y on the other. then, we can simplify to solve for y

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What are the points of intersection of the following functions? F(x)=3x+6 and g(x)=2x^2-3

Dvinal [7]

Answer:

(3,15) and (-1.5, 1.5)

Step-by-step explanation:

If g(x) = 2x² - 3

8 0

3 years ago

A single die is rolled. find the probability of rolling an odd number or a number less than 4.

bonufazy [111]

Odd no. 3/6 = 1/2
less than 4 3/6 = 1/2

6 0

2 years ago

Find the domain and range of y=-5x-1

harina [27]

Answer:

$\mathrm{Domain\:of\:}\:-5x-1\::\quad \begin{bmatrix}\mathrm{Solution:}\:&\:-\infty \:$

$\mathrm{Range\:of\:}-5x-1:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:-\infty \:$

Step-by-step explanation:

Finding the domain:

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

Given the function

$y=-5x-1$

The function has no undefined points nor domain constraints. Hence, the domain is

$-\infty \:$

i.e.

$\mathrm{Domain\:of\:}\:-5x-1\::\quad \begin{bmatrix}\mathrm{Solution:}\:&\:-\infty \:$

Finding the range:

The range of a function is the set of possible output values (dependent variable y values) for which a function is defined.

The range of polynomials with odd degree is all the real numbers.

Hence, the domain is

$-\infty \:$

i.e.

$\mathrm{Range\:of\:}-5x-1:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:-\infty \:$

5 0

2 years ago

Let us have four distinct collinear points $a,$ $b,$ $c,$ and $d$ on the cartesian plane. the point $c$ is such that $\dfrac{ab}

Deffense [45]

Start with a line segment connecting two points, A and B. $\dfrac{DA}{BA}=3$ means DA is 3 times longer than BA. Clearly, D cannot fall between A and B because that would mean DA is shorter than BA. So there are two possible locations where D can be placed on the line relative to A and B.

But with $\dfrac{DB}{BA}=2$ , or the fact that DB is 2 times longer than BA, we can rule out one of these positions; referring to the attachment, if we place D to the left of A, then DB would be 4 times longer than BA.

Finally, $\dfrac{AB}{CB}=\dfrac12$ , so that CB is 2 times longer than AB. Again we have two possible locations for point C (it cannot fall between A and B), but one of them forces C to occupy the same point as D. However, A, B, C, D are distinct, so C must fall to the left of A.

Now let $d$ be the length of AB. Then the length of CD in terms of $d$ is $4d$ . We have the coordinates of C and D, and the distance between them is $\sqrt{(4-0)^2+(0-4)^2}=4\sqrt2$ . So

$4d=4\sqrt2\implies d=\sqrt2$

The slope of the line through C and D is

$\dfrac{0-4}{4-0}=-1$

and so the equation of the line through these points is

$y-4=-(x-0)\implies x+y=4$

So the coordinates of A are $(x,y)=(x,4-x)$ . The distance between C and A is $d=\sqrt2$ , so we have

$\sqrt{(x-0)^2+(4-x-4)^2}=\sqrt{2x^2}=|x|\sqrt2=\sqrt2\implies|x|=1$

Since A falls to the right of C (in the $x,y$ plane, not just in the sketch), we know to take the positive value $x=1$ . Then the $y$ coordinate is $y=4-1=3$ .