All categories

3 years ago

Ten times the quotient of 104 and 88

Mathematics

1 answer:

lilavasa [31]3 years ago

6 0

11.82 which is closest to 12, would be the answer to ten times the quotient of 104 and 88.

You might be interested in

HELP ME PLZ ASAPPPPPP

boyakko [2]

Answer:

$(x,y) =(-4,-3)$ --- Vertex

$x = -4$ --- Axis of symmetry

Step-by-step explanation:

Given

$y = -6(x + 4)^2 - 3$

Solving (a): The vertex

For an equation written in

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

The vertex is:

$(x,y) = (h,k)$

By comparison:

$y = a(x - h)^2 + k$ and $y = -6(x + 4)^2 - 3$

$-h =4$ $k = -3$

$h =-4$ $k = -3$

So, the vertex is:

$(x,y) =(-4,-3)$

Solving (b): The axis of symmetry

For an equation written in

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

The axis of symmetry is:

x = h

In (a):

$h =-4$

So:

$x = -4$

5 0

2 years ago

Find the total area for the regular pyramid T.A. =

Stolb23 [73]

Answer:

The answer to your question is At = 48 + 36 $\sqrt{3}$

Step-by-step explanation:

Area = Area of the base + Lateral area

a) Area of the base

Area = length x height / 2

length = 12

height² = 12² - 6²

= 144 - 36

= 108

height = $\sqrt{108}$ = 6 $\sqrt{3}$

Area = 12 x 6 $\sqrt{3} / 2$

= 36 $\sqrt{3}$

b) Lateral area

height² = 10² - 6²

= 100 - 36

= 64

height = 8

Area = 12 x 8/2

= 48

c) Total area

At = 48 + 36 $\sqrt{3}$

4 0

2 years ago

If there are 6 hours in a school day, 5 school days in a week, 4 school weeks in a month and 9 months in a school year, how many

tino4ka555 [31]

Answer:

1080

Step-by-step explanation:

6 0

2 years ago

How do you math please help me

Stella [2.4K]

Math is math math gives more math math is math math is more math more math is tons of math tons of math is big brain math knowledge big brain math knowledge means your math

3 0

3 years ago

Suppose a population grows according to the logistic equation but is subject to a constant total harvest rate of H. If N(t) is t

Elenna [48]

Answer:

a) Equilibrium point : [ 947, 53 ]

b) N = 947 is stable equilibrium, N = 53 is unstable equilibrium

c) N0, the population will not go extinct

Step-by-step explanation:

Given that;

r = 2, k = 1000, H = 100

dN/dT = R(1 - N/k)N - H

so we substitute

dN/dt = 2( 1 - N/1000)N - 100

now for equilibrium solution, dN/dt = 0

2( 1 - N/1000)N - 100 = 0

((1000 - N)/1000)N = 50

N^2 - 1000N + 50000 = 0

N = 1000 ± √(-1000)² - 4(1)(50000)) / 2(1)

N = 947.213 OR 52.786

approximately

N = 947 OR 53

Therefore Equilibrium point : [ 947, 53 ]

g(N) = 2( 1 - N/1000)N - 100

= 2N - N²/500 - 100

g'(N) = 2 - N/250

SO AT 947

g'(N) = g'(947) = 2 - 947/250 = -1.788 which is less than (<) 0

so N = 947 is stable equilibrium

now AT 53

g'(N) = g"(53) = 2 - 53/250 = 1.788 which is greater than (>) 0

so N = 53 is unstable equilibrium

The capacity k=1000

If the population is less than 53 then the population will become extinct but since the capacity is equal to 1000 then the population will not go extinct.

6 0

3 years ago