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2 years ago

9

Please help me solve this question will give you braisnlt !

1 answer:

Effectus [21]2 years ago

7 0

I believe it’s letter A tho no cap

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Hellllppppp meeeeeee plzzzzz fasttttt

valentinak56 [21]

Answer:

A rate: a measure, quantity, or frequency, typically one measured against some other quantity or measure.

A unit rate: When rates are expressed as a quantity of 1, such as 2 feet per second or 5 miles per hour, they are called unit rates. If you have a multiple-unit rate such as 120 students for every 3 buses, and want to find the single-unit rate, write a ratio equal to the multiple-unit rate with 1 as the second term.

Hope this helps.

4 0

3 years ago

A scientist measure change in the temperature of chemical solution at the start of the experiment , the solutions is at temperat

zhannawk [14.2K]

Answer:

The temperature of the chemical solution after t hours is $-14.96-2.9t\;^{\circ}C$

Step-by-step explanation:

Given that the initial temperature of the chemical solution, $T_0= -14.96^{\circ}C$ .

The rate of decreasing temperature, $R = 2.9^{\circ}C/ hour$

So, after t hours, the temperature of the chemical solution will decrease by $Rt=2.9t^{\circ}C$ .

So, the temperature of the chemical solution after t hours,

$T=T_0-Rt \\\\\Rightarrow T = -14.96-2.9t\;^{\circ}C$

Hence, the temperature of the chemical solution after t hours is $-14.96-2.9t^{\circ}C$ .

4 0

2 years ago

4. Express each of the following in m/s.

Alex17521 [72]

${ \large{ \purple{ \sf{Hello \: there!}}}}$

As we know that, 1 cm = 0.01 m

So,

(a) 65 cm/s = 0.65 m/s

b) 367 km/h = (367 × 5 ÷ 18) m/s

c) 1000 cm/min = (1000 × 60 ÷ 100) m/s

d) 86 km/min = (86 × 60 ÷ 1000) m/s

$\star \:{ \underline{ \boxed{ \sf{ \blue{Thank \: you : ) }}}}}$

8 0

3 years ago

The area of a sheet of paper is approximately

ElenaW [278]

Answer:

1/2²m²

Step-by-step explanation:

7 0

3 years ago

Which table contains the values that satisfy equation y=0.5×+12

alina1380 [7]

Answer: 6

Step-by-step explanation:

Maybe

6 0

2 years ago