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ale4655 [162]
2 years ago
6

What is the y-coordinate of the plotted point?

Mathematics
2 answers:
Eddi Din [679]2 years ago
7 0
The y-coordinate of the ordered pair, (3,2) is 2.
Nana76 [90]2 years ago
3 0

Answer:

(3,2)

Step-by-step explanation:

run before you jump

You might be interested in
What is the expression of (5)(x+3)(y+2)
Elanso [62]

Answer:

The expression of (5)(x+3)(y+3) is a binomial

5 0
2 years ago
At what value of x does the graph of the following function f(x) have a vertical asymptote? f(x)=1/x+3
ratelena [41]

Answer:

d

Step-by-step explanation:

The denominator of f(x) cannot be zero as this would make f(x) undefined.

Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non zero for this value then it is a vertical asymptote.

x + 3 = 0 ⇒ x = - 3 ← equation of vertical asymptote

5 0
3 years ago
Read 2 more answers
PLEASE HELP ME . how can the converse of the Pythagorean theorem help you determine whether the roped off area is in the shape o
trapecia [35]

Step-by-step explanation:

From the Pythagorean theorem we have:

a^2+b^2=c^2           <em>subtract b² from both sides</em>

a^2=c^2-b^2\to a=\sqrt{c^2-b^2}

Same:

b=\sqrt{c^2-a^2}

4 0
3 years ago
The graph of a quadratic function contains (4,-64) and it's vertex is at the origin.write an equation for this function.
lesya692 [45]

Answer:

Step-by-step explanation:

Not sure what form you need this in, but it really doesn't matter, as you'll see in the final equation.  I used the vertex form and solved for a:

y=a(x-h)^2+k

We are given the vertex (h, k) as the origin (0, 0), and we have a point that the graph goes through as (4, -64).  That's our x and y.  Plugging in what we have:

-64=a(4-0)^2+0 gives us

-64 = 16a and

a = -4.  That means that the quadratic equation is

y=-4x^2  which is both vertex form and standard form here, no difference.

5 0
3 years ago
PLEASE HELP NOW! I WILL MARK BRAINLIEST!
Natasha_Volkova [10]

Answer:

17 units

Step-by-step explanation:

Given 2 congruent chords in the circle, then they are equidistant from the centre and perpendicular

There is a right triangle formed by legs 15 and 8, with radius r being the hypotenuse.

Using Pythagoras' identity in this right angle

r² = 15² + 8² = 225 + 64 = 289 ( take the square root of both sides )

r = \sqrt{289} = 17

7 0
2 years ago
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