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4 years ago

The temperature dropped 24 degrees in 4 hours. Which integer represents the change per hour?

Mathematics

1 answer:

Lilit [14]4 years ago

7 0

Answer:

-6

Step-by-step explanation:

24/4 is 6 but the tempature decreased due to that we put a negative symbol in front.

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1 point

Evgesh-ka [11]

Answer:

Its an integer and rational number

7 0

3 years ago

Sarah sells fruit trays at a salad shop. Last week, she sold 450 fruit trays for $8 per tray. In previous weeks, she found that

Mamont248 [21]

Sarah's income = Number of trays × Rate per tray

y = 450 × 8

If there is an increase of 0.85 in price,

New price = 8 + 0.85

For x number of increases,

new price = 8 + 0.85x.

Number of trays reduced = 450 - 15x.

So, Sarah's income in this case is y = (450 - 15x) (8 + 0.85x).

4 0

4 years ago

Read 2 more answers

10 ft

nekit [7.7K]

10 x 8 = 80ft for the rectangle
To work out the area of the semi circle you do pi (3.14) x the radius squared, the radius is half the diameter.
3.14 x 4 squared
3.14 x16 = 50.24ftsqaured for the whole circle so you divide it by 2 which is 25.12ft squared.
And then finally you add 25.12 and 80 so your answer is 105.12 ft squared

5 0

3 years ago

A student stands on a bathroom scale in an elevator at rest on the 64 floor of a building. the scale reads 836n. as the elevator

strojnjashka [21]

At rest
mg=836 N
as g=9.8
m=836/9.8=85.30 kg
when going up,
mg+ma=935N
ma=935-836= 99N

a=99/85.30 = 1.16 m/s².
that is the acceleration of elevator
pls marks as brainliest answer

3 0

3 years ago

Let f(x) = (x − 1)2, g(x) = e−2x, and h(x) = 1 + ln(1 − 2x). (a) Find the linearizations of f, g, and h at a = 0. What do you no

sweet [91]

Answer:

Lf(x) = Lg(x) = Lh(x) = 1 - 2x

value of the functions and their derivative are the same at x = 0

Step-by-step explanation:

Given :

f(x) = (x − 1)^2,

g(x) = e^−2x ,

h(x) = 1 + ln(1 − 2x).

a) Determine Linearization of f, g and h at a = 0

L(x) = f (a) + f'(a) (x-a) ( linearization of f at a )

for f(x) = (x − 1)^2

f'(x ) = 2( x - 1 )

at x = 0

f' = -2

hence the Linearization at a = 0

Lf (x) = f(0) + f'(0) ( x - 0 )

Lf (x) = 1 -2 ( x - 0 ) = 1 - 2x

For g(x) = e^−2x

g'(x) = -2e^-2x

at x = 0

g(0) = 1

g'(0) = -2e^0 = -2

hence linearization at a = 0

Lg(x) = g ( 0 ) + g' (0) (x - 0 )

Lg(x) = 1 - 2x

For h(x) = 1 + ln(1 − 2x).

h'(x) = -2 / ( 1 - 2x )

at x = 0

h(0) = 1

h'(0) = -2

hence linearization at a = 0

Lh(x) = h(0) + h'(0) (x-0)

= 1 - 2x

Observation and reason

The Linearization is the same in every function i.e. Lf(x) = Lg(x) = Lh(x) this is because the value of the functions and their derivative are the same at x = 0

8 0

3 years ago