What is 4/5 x for x =5/8? HALP pls

Does anyone know how to evaluate this problem? im stuck

shusha [124]

If you are just supposed to evaluate the a) f(0) and b) f(4) then you would just plug it in the equation.
f(0)=0^2 - 8*0 + 16
f(0)=0 - 0 + 16
f(0)=16
f(4)=4^2 - 8*4 +16
f(4)=16 - 32 + 16
f(4)=0
Feel free to comment any questions and if there is any additional context to the question please say so I can help with that.

5 0

3 years ago

Simplify 27x^7y^2/9x^6y^5

solmaris [256]

Answer: x/y^3

Step-by-step explanation:

So you divide 27/9 which is 3. Then it ix x^7-x^6=x. And then it is y^2-y^5=y3

27*x*x*x*x*x*x*x*y*y/9*x*x*x*x*x*x*y*y*y*y*y

Doing this makes it easier so you cross each variable out for each one in both sides. 3x/y^3 would be your answer.

7 0

3 years ago

Please help solve with work

sergey [27]

X-4y=28
8x+4y=8
---------------
9x=36→x=4
4-4y=28→y=6
—-----------------------------------------------------------
y-5=x
4(y-5)-y=4
4y-20-y=4
3y=24
y=8
x8-5=3

4 0

3 years ago

Identify the vertex of the function, F(x) =3(x-1)2 + 5.

agasfer [191]

Answer:

f

(

x

)

=

3

(

x

−

1

)

2

+

5

Use the vertex form,

y

=

a

(

x

−

h

)

2

+

k

, to determine the values of

a

,

h

, and

k

.

a

=

3

h

=

1

k

=

5

Find the vertex

(

h

,

k

)

.

(

1

,

5

)

image of graph

8 0

2 years ago

An airplane travels 6111 kilometers against the wind in 9 hours and 7911 kilometers with the wind in the same amount of time. Wh

natima [27]

Answer:

Speed of the plane in still air: $779\; {\rm km \cdot h^{-1}}$ .

Windspeed: $100\; {\rm km \cdot h^{-1}}$ .

Step-by-step explanation:

Assume that $x\; {\rm km \cdot h^{-1}}$ is the speed of the plane in still air, and that $y\; {\rm km \cdot h^{-1}}$ is the speed of the wind.

When the plane is travelling against wind, the ground speed of this plane (speed of the plane relative to the ground) would be $(x - y)\; {\rm km \cdot h^{-1}}$ .
When this plane is travelling in the same direction as the wind, the ground speed of this plane would be $(x + y)\; {\rm km \cdot h^{-1}}$ .

The question states that when going against the wind ( $v = (x - y)\; {\rm km \cdot h^{-1}}$ ,) the plane travels $6111\; {\rm km}$ in $9\; {\rm h}$ . Hence, $9\, (x - y) = 6111$ .

Similarly, since the plane travels $7911\; {\rm km}$ in $9\; {\rm h}$ when travelling in the same direction as the wind ( $v = (x + y)\; {\rm km \cdot h^{-1}}$ ,) $9\, (x + y) = 7911$ .

Add the two equations to eliminate $y$ . Subtract the second equation from the first to eliminate $x$ . Solve this system of equations for $x$ and $y$ : $x = 779$ and $y = 100$ .

Hence, the speed of this plane in still air would be $779\; {\rm km \cdot h^{-1}}$ , whereas the speed of the wind would be $100\; {\rm km \cdot h^{-1}}$ .

3 0

1 year ago