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2 years ago

(6,0) (0,2) (3,8) (5,5)

Mathematics

1 answer:

Luden [163]2 years ago

3 0

Answer:

what is the question you need help with??

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Find the zeros of the function y=x^2-1

Sergio [31]

Answer:

-1 and 1 are the zeros

Step-by-step explanation:

Graph it and youll see that the curve touches the x axis at (-1,0) and (1,0)

5 0

3 years ago

Read 2 more answers

Find k so that the following function is continuous: f(x)={kx8x2if0≤x<5if5≤x.

tankabanditka [31]

Check the one-sided limits:

$\displaystyle \lim_{x\to5^-}f(x) = \lim_{x\to5}kx = 5k$

$\displaystyle \lim_{x\to5^+}f(x) = \lim_{x\to5}8x^2 = 200$

If f(x) is to be continuous at x = 5, then these two limits should have the same value, which means

5k = 200

k = 200/5

k = 40

3 0

2 years ago

Ryleigh received a $50 gift card to the frozen yogurt shop for her birthday. The shop sells yogurt sundaes for $4 and yogurt con

timama [110]

The answer is D because the shaded area must go to the origin and it must have a non-dashed line because having 5 sundaes and 10 cones is possible.

9 0

3 years ago

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Determine whether f(x) = –5x^2 – 10x + 6 has a maximum or a minimum value.

Blizzard [7]

First,

We are dealing with parabola since the equation has a form of,

$y=ax^2+bx+c$

Here the vertex of an up - down facing parabola has a form of,

$x_v=-\dfrac{b}{2a}$

The parameters we have are,

$a=-5,b=-10, c=6$

Plug them in vertex formula,

$x_v=-\dfrac{-10}{2(-5)}=-1$

Plug in the $x_v$ into the equation,

$y_v=-5(-1)^2-10(-1)+6=11$

We now got a point parabola vertex with coordinates,

$(x_v, y_v)\Longrightarrow(-1,11)$

From here we emerge two rules:

If $a$ then vertex is max value
If $a>0$ then vertex is min value

So our vertex is minimum value since,

$a=-5\Longleftrightarrow a$

Hope this helps.

r3t40

7 0

3 years ago

Please answer! Will give brainliest please and thank you

nydimaria [60]

Answer:

119 square ft is the answer now give me brainliest

8 0

2 years ago