Ask question

Ask question

All categories

stealth61 [152]

3 years ago

14

H(x) = 3x + 5 g(x) = 4x – 2 Find h(x) – g(x)

1 answer:

Flura [38]3 years ago

8 0

Answer:

6

Step-by-step explanation:

seshhhhh is that dababy. he valid

You might be interested in

If a ÷ b = 2 and b ÷ c =3/4, what is the value of c ÷ a? Express your answer as a common fraction.

lora16 [44]

Answer:

in the picture above.

Step-by-step explanation:

I hope that it's a correct answer.

7 0

3 years ago

A physics student stands at the top of a hill that has an elevation of 56 meters. He throws a rock and it goes up into the air a

Virty [35]

round your answer to the nearest hundredth place the answer is a

3 0

3 years ago

The height of a stuntperson jumping off a building that is 20 m high is modeled by the equation h = 20 – 5t2, where t is the tim

Nostrana [21]

Answer:

$\sqrt{2}$ - 1

Step-by-step explanation:

h = 20 – 5t2 = 15 -> t = 1

h = 20 – 5t2 = 10 -> t = $\sqrt{2}$

interval $\sqrt{2}$ - 1

8 0

3 years ago

Read 2 more answers

A recipe uses 5 cups of flour for every 2 cups of sugar. How much sugar is used for 1 cup of flour? Exact answers only

Aloiza [94]

Answer:

0.4

Step-by-step explanation:

Hopefully this can help

7 0

3 years ago

EXAMPLE 5 Find the maximum value of the function f(x, y, z) = x + 2y + 9z on the curve of intersection of the plane x − y + z =

geniusboy [140]

The Lagrangian,

$L(x,y,z,\lambda,\mu)=x+2y+9z-\lambda(x-y+z-1)-\mu(x^2+y^2-1)$

has critical points where its partial derivatives vanish:

$L_x=1-\lambda-2\mu x=0$

$L_y=2+\lambda-2\mu y=0$

$L_z=9-\lambda=0$

$L_\lambda=x-y+z-1=0$

$L_\mu=x^2+y^2-1=0$

$L_z=0$ tells us $\lambda=9$ , so that

$L_x=0\implies-8-2\mu x=0\implies x=-\dfrac4\mu$

$L_y=0\implies11-2\mu y=0\implies y=\dfrac{11}{2\mu}$

Then with $L_\mu=0$ , we get

$x^2+y^2=\dfrac{16}{\mu^2}+\dfrac{121}{4\mu^2}=1\implies\mu=\pm\dfrac{\sqrt{185}}2$

and $L_\lambda=0$ tells us

$x-y+z=-\dfrac4\mu-\dfrac{11}{2\mu}+z=1\implies z=1+\dfrac{19}{2\mu}$

Then there are two critical points, $\left(\pm\frac8{\sqrt{185}},\mp\frac{11}{\sqrt{185}},1\pm\frac{19}{\sqrt{185}}\right)$ . The critical point with the negative $x$ -coordinates gives the maximum value, $9+\sqrt{185}$ .

8 0

4 years ago