Ask question

Ask question

All categories

3 years ago

10

If f(x) = 72(1.25)^x and f(1) = 72(1.25)^1 = 11.25 f(2) = 72(1.25)^2 = ?????????

1 answer:

KengaRu [80]3 years ago

6 0

Answer:

f(1)=101.25 and f(2)=112.5

Step-by-step explanation:

f(1) = 72(1.25)^1 = 11.25

f(1)=90=11.25

f(1)=101.25

f(2) = 72(1.25)^2

f(2)=72(1.5625)

f(2)=112.5

You might be interested in

5) h(x) = 2x + 4; Find h(-3)

olya-2409 [2.1K]

Answer:

-2

Step-by-step explanation:

You make x= -3 then you insert it into the problem which is x= 2(-3)+4

2 times -3 is -6 which then you would get

-6+4 which equals -2

5 0

3 years ago

Explain how to use the missing pieces strategy to compare two fractions.include a diagram with your explanation.

umka21 [38]

In order for us to tell whichever fraction is greater than or lesser than, we need to look for the missing part.

For example, we have 2/3 and 1/3.
2/3 needs to have 1/3 in order to complete the whole.
1/3 needs to have 2/3 in order to complete the whole.

1/3 < 2/3
<span>So 2/3 is bigger than 1//3</span>

7 0

4 years ago

PLEASE GUYS HELP ME I NEED THIS FOR A TEST NOW

ValentinkaMS [17]

Answer:

23.54 m

Step-by-step explanation:

Applying

cos∅ = adjacent(A)/hypotenuse(H)

cos∅ = A/H................ Equation 1

make H the subject of the equation

H = A/cos∅............ Equation 2

Given: A = 15 m, ∅ = 25°

Substitute into equation 2

H = 15/cos25

H = 16.55 m

Also,

tan∅ = opposite(O)/Adjacent(A)

tan∅ = O/A............Equation 3

Make O the subject of the equation

O = Atan∅.......... Equation 4

Substituting into equation 4

O = 15(tan25°)

O = 6.99 m.

From the diagram,

The height of the goal post before snap = H+O

The height of the goal post before snap = 16.55+6.99

The height of the goal post before snap = 23.54 m

4 0

3 years ago

Consider F and C below. F(x, y, z) = yz i + xz j + (xy + 6z) k C is the line segment from (2, 0, −2) to (5, 4, 2) (a) Find a fun

WITCHER [35]

Answer:

Required solution (a) $f(x,y,z)=xyz+3z^2+C$ (b) 40.

Step-by-step explanation:

Given,

$F(x,y,z)=yz \uvec i +xz\uvec j+(xy+6z)\uvex k$

(a) Let,

$F(x,y,z)=yz \uvec i +xz\uvec j+(xy+6z)\uvex k=f_x \uvec{i} +f_y \uvex{j}+f_z\uvec{k}$

Then,

$f_x=yz,f_y=xz,f_z=xy+6z$

Integrating $f_x$ we get,

$f(x,y,z)=xyz+g(y,z)$

Differentiate this with respect to y we get,

$f_y=xz+g'(y,z)$

compairinfg with $f_y=xz$ of the given function we get,

$g'(y,z)=0\implies g(y,z)=0+h(z)\implies g(y,z)=h(z)$

Then,

$f(x,y,z)=xyz+h(z)$

Again differentiate with respect to z we get,

$f_z=xy+h'(z)=xy+6z$

on compairing we get,

$h'(z)=6z\implies h(z)=3z^2+C$ (By integrating h'(z)) where C is integration constant. Hence,

$f(x,y,z)=xyz+3z^2+C$

(b) Next, to find the itegration,

$\int_C \vec{F}.dr=\int_C \nabla f. d\vec{r}=f(5,4,2)-f(2,0,-2)=(52+C)-(12+C)=40$

3 0

3 years ago

PLA HELP QUESTION IN PHOTO

asambeis [7]

Answer:

Step-by-step explanation:

its the first one

7 0

3 years ago

Read 2 more answers