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3 years ago

Help with question 2 please.

Mathematics

1 answer:

Firdavs [7]3 years ago

3 0

Words would be "x is smaller than 1"

Algebra would be x < 1

Possible solutions are 0, -1, -2

Hope this helps.

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find the area of a segment of a circle that subtends an angle of 120° at the centre of a circle of radius 56cm.

Kazeer [188]

Answer:

Area = 1,527 in

Step-by-step explanation:

Circle area = π * r² = π * 3136 [cm²] ≈ 1527 [in²]

π ≈ 3.14159265 ≈ 3.14

d = r * 2 = 56 [cm] * 2 = 44.094 [inch]

8 0

2 years ago

Two ships are sailing parallel to each other. The path of one ship is represented on a coordinate plane as the line y = –13x + 4

wlad13 [49]

Answer:

yo i would help but my brain jus farted jus by lookin at it

Step-by-step explanation:

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3 years ago

At 2 PM, a thermometer reading 80°F is taken outside where the air temperature is 20°F. At 2:03 PM, the temperature reading yiel

Alexeev081 [22]

Use Law of Cooling:
$T(t) = (T_0 - T_A)e^{-kt} +T_A$
T0 = initial temperature, TA = ambient or final temperature

First solve for k using given info, T(3) = 42
$42 = (80-20)e^{-3k} +20 \\ \\ e^{-3k} = \frac{22}{60}=\frac{11}{30} \\ \\ k = -\frac{1}{3} ln (\frac{11}{30})$

Substituting k back into cooling equation gives:
$T(t) = 60(\frac{11}{30})^{t/3} + 20$

At some time "t", it is brought back inside at temperature "x".
We know that temperature goes back up to 71 at 2:10 so the time it is inside is 10-t, where t is time that it had been outside.
The new cooling equation for when its back inside is:
$T(t) = (x-80)(\frac{11}{30})^{t/3} + 80 \\ \\ 71 = (x-80)(\frac{11}{30})^{\frac{10-t}{3}} + 80$
Solve for x:
$x = -9(\frac{11}{30})^{\frac{t-10}{3}} + 80$
Sub back into original cooling equation, x = T(t)
$-9(\frac{11}{30})^{\frac{t-10}{3}} + 80 = 60 (\frac{11}{30})^{t/3} +20$
Solve for t:
$60 (\frac{11}{30})^{t/3} +9(\frac{11}{30})^{\frac{-10}{3}}(\frac{11}{30})^{t/3} = 60 \\ \\ (\frac{11}{30})^{t/3} = \frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}} \\ \\ \\ t = 3(\frac{ln(\frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}})}{ln (\frac{11}{30})}) \\ \\ \\ t = 4.959$

This means the exact time it was brought indoors was about 2.5 seconds before 2:05 PM

8 0

3 years ago

Iron can exist as a solid or liquid. Which statement best comparison solid iron and liquid iron?

tester [92]

Answer: Iron, a solid at room temperature, becomes a liquid at 2800 degrees Fahrenheit (really, REALLY hot) and a gas at 5182 degrees Fahrenheit (about half the temperature of the sun). When things are hot, the molecules and atoms move around more and faster, and when they are cold they are slower.

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

AB = 3 + x

ICE Princess25 [194]

The given quadrilateral ABCD is a parallelogram since the opposite sides are of same length AB and DC is 4 and AD and BC is 2.

Step-by-step explanation:

ABCD is a quadrilateral with their opposite sides are congruent (equal).

The both pairs of opposite sides are given as AB = 3 + x , DC = 4x , AD = y + 1 , BC = 2y.

AB and DC are opposite sides and have same measure of length.
AD and BC are opposite sides and have same measure of length.

To find the length of AB and DC :

AB = DC

3 + x = 4x

Keep x terms on one side and constant on other side.

3 = 4x - x

3 = 3x

x = 1

Substiute x=1 in AB and DC,

AB = 3+1 = 4

DC = 4(1) = 4

To find the length of AD and BC :

AD = BC

y + 1 = 2y

Keep y terms on one side and constant on other side.

2y-y = 1

y = 1

Substiute y=1 in AD and BC,

AD = 1+1 = 2

BC = 2(1) = 2

Therefore, the opposite sides are of same length AB and DC is 4 and AD and BC is 2. The given quadrilateral ABCD is a parallelogram.

6 0

3 years ago