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3 years ago

I need help!!! Idk what the answer is

Mathematics

1 answer:

yanalaym [24]3 years ago

3 0

Translations, reflections, and rotations preserve congruency. Dilations do not?

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A line is drawn through (–4, 3) and (4, 3). Which describes whether or not the line represents a direct variation? The line repr

Dovator [93]

Answer:

Line does not represent direct variation.

Step-by-step explanation:

If the line goes thru the origin, it represents direct variation. Otherwise it does not.

Start by finding the slope of this line:

As we move from (-4,3) to (4,3), x increases by 8 but y does not change. Thus, the slope is 0/8, or just 0. Thus, the equation of this line is y = 3, which does NOT go thru the origin; thus, the points given do NOT represent direct variation.

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3 years ago

One survey estimates that, on average, the retail value of a mid-sized car decreases by 8% annually. If the retail value of a ca

ivolga24 [154]

Value in 1 year = V*0.92

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3 years ago

Help ASAP Science/ Math..... Bill and Amy want to ride their bikes from their neighborhood to school which is 14,400 meters away

Oksana_A [137]

Answer:

Amy is 1200 seconds fater than Bill.

Step-by-step explanation:

4 0

3 years ago

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

Can someone help me? please

Black_prince [1.1K]

Answer:

I can but not willing right now so sorry

4 0

3 years ago