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3 years ago

Which correctly applies the distributive property to show an equivalent expression to (8.6)(–2.5)?

2 answers:

7 0

We see all of them have 2 and 0.5
so therfor b+c=-2.5

so

remember
a(b+c)=ab+ac
so
8.6(-2-0.5)=8.6(-2)+8.6(-0.5)

last option

elixir [45]3 years ago

6 0

Answer:

$(8.6)(-2.5)=(8.6)(-2)+(8.6)(-0.5)$

Step-by-step explanation:

Given the expression

$(8.6)(-2.5)$

we have to apply the distributive property

we can write 2.5 as 2+0.5

$(8.6)(-2.5)=(8.6)(-(2+0.5))=(8.6)((-2)+(-0.5))$

By distributive property,

$a.(b+c)=a.b+a.c$

$(8.6).((-2)+(-0.5))=(8.6)(-2)+(8.6)(-0.5)$

Hence, $(8.6)(-2.5)=(8.6)(-2)+(8.6)(-0.5)$

Therefore, the correct option is D

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The base of an aquarium with given volume V is made of slate and the sides are made of glass. If slate costs five times as much

Y_Kistochka [10]

Answer:

x = ∛ 2*V/5

y = ∛ 2*V/5

h = V/ ∛ 4*V²/25

Step-by-step explanation:

Dimensions of the aquarium base is x*y

We call c₁ cost per unit area of the sides, then cost per unit area of slate is equal 5c₁.

let call h the height of the aquarium then volume of the aquarium is:

V = x*y*h where h = V / x*y

As the base is a rectangular one there are 2 sides x*h . and 2 sides y*h

According to this:

Ct (cost of aquarium ) = cost of the base + cost of the sides

cₐ ( cost of the base) = 5*c₁*x*y

c₆ (cost of the sides ) = c₁*2*x*h + c₁*2*y*h

C(t) = 5*c₁*x*y +2* c₁*x* V/x*y + 2* c₁*y* V/x*y or

C(t) = 5*c₁*x*y + 2*c₁*V/y *2*c₁* V/x

Taking partial derivatives en x and y we have:

C´(x) = 5*c₁*y - 2*c₁*V/x²

C´(y) = 5*c₁*x - 2*c₁*V/y²

C´(x) = C´(y) ⇒ 5*c₁*y - 2*c₁*V/x² = 5*c₁*x - 2*c₁*V/y²

or 5*y - 2*V/x² = 5*x - 2*V/y²

(5*y*x² - 2*V)/x² = ( 5*y²x - 2*V) /y²

(5*y*x² - 2*V)*y² = ( 5*y²x - 2*V)*x²

5*y³*x² - 2*V*y² = 5*y²x³ - 2*V*x²

5*y³*x² - 5*y²x³ = 2*V * ( y² - x²)

by symmetry x = y

Then using x = y and plugging that value on the derivatives

C´(x) = 5*c₁*y - 2*c₁*V/x²

C´(x) = 5*c₁*x - 2*c₁*V/x²

C´(x) = 0 ⇒ 5*c₁*x - 2*c₁*V/x² = 0

5*x - 2*V/x² = 0 ⇒ 5*x³ - 2*V = 0 ⇒ 5*x³ = 2*V ⇒ x³ = 2*V/5

x = ∛ 2*V/5 and y = ∛ 2*V/5 and h = V/ x*y h = V/ ∛ 4*V²/25

7 0

3 years ago

Sam drove 320 miles in 2.5 hours, how long will it take to drive 500 at the same rate (speed)?

marishachu [46]

Answer:

D = (55)(4.5)

425 = 60 T

200 = R (2.5)

4.6 hours

59 mph

217 miles

Step-by-step explanation:

6 0

3 years ago

How many yards will the players have run ?? Please help

Whitepunk [10]

Answer:

125.3 yards

Step-by-step explanation:

He wants them to run the so 110^2+60^2 =which is 125.3.

Hope this helps plz mark brainliest :D

8 0

3 years ago

What is the length of the arc on a circle with radius 20 in. intercepted by a 15° angle?

AnnyKZ [126]

Length of arc = (15/360) * 2*3.14*20 = 5.23 ins

circumference of the circle = 2*3.14*24 = 150.72

angle at center from 10 ft arc = (10/150.72) * 360 = 23.89 degrees

6 0

3 years ago

Read 2 more answers

According to data from a medical association, the rate of change in the number of hospital outpatient visits, in millions, in

GaryK [48]

Answer:

a) $f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+264,034,000$

b) f(t=2015) = 264,034,317.7

Step-by-step explanation:

The rate of change in the number of hospital outpatient visits, in millions, is given by:

$f'(t)=0.001155t(t-1980)^{0.5}$

a) To find the function f(t) you integrate f(t):

$\int \frac{df(t)}{dt}dt=f(t)=\int [0.001155t(t-1980)^{0.5}]dt$

To solve the integral you use:

$\int udv=uv-\int vdu\\\\u=t\\\\du=dt\\\\dv=(t-1980)^{1/2}dt\\\\v=\frac{2}{3}(t-1980)^{3/2}$

Next, you replace in the integral:

$\int t(t-1980)^{1/2}=t(\frac{2}{3}(t-1980)^{3/2})- \frac{2}{3}\int(t-1980)^{3/2}dt\\\\= \frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}+C$

Then, the function f(t) is:

$f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+C'$

The value of C' is deduced by the information of the exercise. For t=0 there were 264,034,000 outpatient visits.

Hence C' = 264,034,000

The function is:

$f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+264,034,000$

b) For t = 2015 you have:

$f(t=2015)=0.001155[\frac{2}{3}(2015)(2015-1980)^{1/2}-\frac{4}{15}(2015-1980)^{5/2}]+264,034,000\\\\f(t=2015)=264,034,317.7$

3 0

3 years ago