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1 year ago

What is the approximate value of the function at x = 6?

Mathematics

1 answer:

Gelneren [198K]1 year ago

5 0

Answer:

about 3/4

Step-by-step explanation:

To tell what the y value of a certain x value will be is very simple if we have a graph--you can just look at the portion of the graph where x = 6. At that point, it appears to be 3/4 ish on the y-axis

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Write an equation in point slope form for each line Slope is 3 and (-2,-1) is on the line

elena-14-01-66 [18.8K]

Answer:

$\huge \blue{y + 1 = 3(x + 2) } \\$

Step-by-step explanation:

To find an equation of a line in point slope form when given the slope and a point we use the formula

$y - y_1 = m(x - x_1)$

From the question we have the final answer as

$y + 1 = 3(x + 2)$

Hope this helps you

6 0

3 years ago

Help me please!! it's due today

Bess [88]

Answer:

the answer would be 14-16=-2

Step-by-step explanation:

first u look where the first long arrow goes on this problem then you follow the second arrow which stops at 6 then u count how much until negative 2. but how I did it is the form 14 how much to 0 which is 14 the I count how many until I get to -2 and there's the answer

3 0

2 years ago

($23 dollars per person)The cost of the hall for the dance is $4000 dollars How many people must attend the dance to cover the r

garik1379 [7]

Answer:

174

Step-by-step explanation:

Rental Fee ÷ Cost of Ticket = # of people needed

$4000÷$23=173.9....

Can't have .9 of a ticket, so round up to 174.

# of people needed is 174

8 0

2 years ago

I need some help with math

Irina18 [472]

Answer:

$(y-7)^2+(x-2)^2=16$

and

$(x+2)^2+(y-15)^2 = 9$

Step-by-step explanation:

The standard equation of a circle is $(x-h)^2+(y-k)^2=r^2$ where the coordinate (h,k) is the center of the circle.

Second Problem:

We can start with the second problem which uses this info very easily.
(h,k) in this problem is (-2,15) simply plug these into the equation. $(x--2)^2+(y-15)^2=r^2$ .
We can also add the radius 3 and square it so it becomes 9. The equation.
This simplifies to $(x+2)^2+(y-15)^2 = 9$ .

First Problem:

The first problem takes a different approach it is not in standard form. But we can convert it to standard form by completing the square.
$y^2-14y+x^2-4x+37=0$ first subtract 37 from both sides so the equation is now $y^2-14y+x^2-4x=-37$ .
$y^2-14y+x^2-4x+37=0$ by adding $(-\frac{b}{2a} )^2$ to both the x and y portions of this equation you can complete the squares. $(-\frac{b}{2a})^2=(-\frac{-14}{2(1)})^2$ and $(-\frac{-4}{2(1)})^2$ which equals 49 and 4.
Add 49 and 4 to both sides and the equation is now: $y^2-14y+49+x^2-4x+4=-37+49+4$ You can simplify the y and x portions of the equations into the perfect squares or factored form $(y-7)^2$ and $(x-2)^2$ .
Finally put the whole thing together. $(y-7)^2+(x-2)^2=16$ .