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4 years ago

9

Rosa has a garden divided into sections. She has 125 daisy plants. If she plants in each section of daisies, will she have any l

eft over? If so, how many daisy plants will be left over?

1 answer:

SashulF [63]4 years ago

8 0

How many sections are there?

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Find f(-11/5) if f(n) = 5n + 6. -17 5 -5

Ber [7]

F(-11/5) if f(n) = 5n + 6. f (-11/5) = 5 (-11/5) + 6 f (-11/5) = -11 + 6 f(-11/5) = -5 :)

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Which number(s) below belong to the solution set of the inequality Check all that apply. x + 30 < 60 a.50 b.45 c.300 d.1 e.29

Igoryamba

Answer:

e.29 d.1

Step-by-step explanation:

x+30<60 :

we will replace with the values we have :

300 that's a big number . Obviously it isn't a solution
50 : 30+50 = 80>60 so no
45: 30+45= 75 >60 so no
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1 yes . it's obvious
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Need help with this maybe question

Aleonysh [2.5K]

Answer:

the vertex is:

(2, -1)

Step-by-step explanation:

First solve the equation for the variable y

$x^2-16y-4x-12=0$

Add 16y on both sides of the equation

$16y=x^2-16y+16y-4x-12$

$16y=x^2-4x-12$

Notice that now the equation has the general form of a parabola

$ax^2 +bx +c$

In this case

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Add $(\frac{b}{2}) ^ 2$ and subtract $(\frac{b}{2}) ^ 2$ on the right side of the equation

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$(\frac{b}{2}) ^ 2=(-2) ^ 2$

$(\frac{b}{2}) ^ 2=4$

$16y=(x^2-4x+4)-4-12$

Factor the expression that is inside the parentheses

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

Divide both sides of the equality between 16

$\frac{16}{16}y=\frac{1}{16}(x-2)^2-\frac{16}{16}$

$y=\frac{1}{16}(x-2)^2-1$

For an equation of the form

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

the vertex is: (h, k)

In this case

$h=2\\k =-1$

the vertex is:

(2, -1)

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