1. Describe one point on the line and the slope of the line graphed below:

Mathematics

1 answer:

PolarNik [594]2 years ago

3 0

The last one :)
Not 100% sure but yeah-

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Coco needs 150mg of a medication that comes in tablets labelled 100mg each. How many tablets will you give her?

shtirl [24]

Answer:

1.5

Step-by-step explanation:

If 1 tablet is 100mg, and you need 150mg, we set up this equation:

Let x be the number of tablets Coco needs.

x = 150/100

x = 1.5

Therefore, you need 1.5 tablets for the medication.

7 0

2 years ago

Dang this is hard quiz

olga nikolaevna [1]

Answer:

-54

Step-by-step explanation:

8 0

2 years ago

Can someone help me would the answer be x -2<br><img src="https://tex.z-dn.net/?f=%20%5Cgeqslant%20" id="TexFormula1" title=" \g

DerKrebs [107]

The answee is -2 < x < 10

6 0

2 years ago

Integration: How would I get to this answer?

Scrat [10]

Given that $y$ attains a maximum at $x=1$ , it follows that $y'=0$ at that same point. So integrating once gives

$\displaystyle\int\frac{\mathrm d^2y}{\mathrm dx^2}\,\mathrm dx=\int-8x\,\mathrm dx$
$\dfrac{\mathrm dy}{\mathrm dx}=-4x^2+C_1$
$\implies -4(1)^2+C_1=0\implies C_1=4$

and so the first derivative is $y'=-4x^2+4$ .

Integrating again, you get

$\displaystyle\int\frac{\mathrm dy}{\mathrm dx}\,\mathrm dx=\int(-4x^2+4)\,\mathrm dx$
$y=-\dfrac43x^3+4x+C_2$

You know that this curve passes through the point (2, -1), which means when $x=2$ , you have $y=-1$ :

$-1=-\dfrac43(2)^3+4(2)+C_2$
$\implies C_2=\dfrac53$

and so

$y=-\dfrac43x^3+4x+\dfrac53$