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3 years ago

Can anyone please help me it’s only one question please help me

Mathematics

1 answer:

Veronika [31]3 years ago

3 0

Answer:

A =7.02 m^2

Step-by-step explanation:

The area of a trapezoid is given by

A = 1/2 (b1+b2)*h where b1 and b2 are the lengths of the bases

A = 1/2 ( 4.5+6.3) * 1.3

A = 1/2(10.8) * 1.3

A =7.02 m^2

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Un jardinier a récolté deux quintaux trois quarts de pommes . Il vend un quintal un quart à un voisin ; il vend aussi 7/10 de qu

Flauer [41]

Answer:

Step-by-step explanation:

Notez que: un quintal =

une unité de poids égale à 100 kg

Un jardinier a récolté deux quintaux et trois quarts de pommes.

Le nombre total de pommes récoltées = 200 kg + 3/4 du quintal de pommes

= 200 kg + 0,381 kg

= 200,381 kg

Il vend un quintal un quart à un voisin

Le montant qu'il a vendu à un voisin = 100 + 1/4 quintal de pommes

100 + 0,127

= 100,127 kg

Il vend également 7/10 d'un quintal sur le marché du village

= 7/10 × 100 kg

= 70 kg de pommes restantes

2/5 d'un quintal à un pâtissier.

2/5 × 100 kg

= 40 kg de pommes restantes

À l'aide d'une fraction, exprimez la quantité de pomme que le jardinier a gardée pour lui.

La quantité que le jardinier lui a réservée =

200 3/4 kg - (100 1/4 kg + 70 kg + 40 kg)

200,381 kg - (210,127 kg)

5 0

3 years ago

What is the midpoint of the line segment that joins (6-2) and (,5)

PIT_PIT [208]

Answer:

Step-by-step explanation:

X=(6,-2)

Y=(0,5)

3 0

2 years ago

I am so lost, please help

skelet666 [1.2K]

Answer:

$u=\frac{9}{t}-\frac{1}{2} a t$

Step-by-step explanation:

Step 1: Given equation: $9=\frac{1}{2} a t^{2}+u t$

To get u, by subtraction equality property subtract both sides of the equation by $\frac{1}{2} a t^{2}$ .

$\Rightarrow 9-\frac{1}{2} a t^{2}=\frac{1}{2} a t^{2}+u t-\frac{1}{2} a t^{2}$

$\Rightarrow 9-\frac{1}{2} a t^{2}=u t$

Step 2: By division equality property, divide both sides of the equation by t.

$\Rightarrow \frac{9-\frac{1}{2} a t^{2}}{t}=\frac{u t}{t}$

$\Rightarrow \frac{9}{t}-\frac{1}{2} a t=u$

Therefore, $u=\frac{9}{t}-\frac{1}{2} a t$ .

So, in Emily’s physics class, she got $u=\frac{9}{t}-\frac{1}{2} a t$ .

7 0

3 years ago

Suppose that a spherical droplet of liquid evaporates at a rate that is proportional to its surface area: where V = volume (mm3

Alex

Answer:

V = 20.2969 mm^3 @ t = 10

r = 1.692 mm @ t = 10

Step-by-step explanation:

The solution to the first order ordinary differential equation:

$\frac{dV}{dt} = -kA$

Using Euler's method

$\frac{dVi}{dt} = -k *4pi*r^2_{i} = -k *4pi*(\frac {3 V_{i} }{4pi})^(2/3)\\ V_{i+1} = V'_{i} *h + V_{i} \\$

Where initial droplet volume is:

$V(0) = \frac{4pi}{3} * r(0)^3 = \frac{4pi}{3} * 2.5^3 = 65.45 mm^3$

Hence, the iterative solution will be as next:

i = 1, ti = 0, Vi = 65.45

$V'_{i} = -k *4pi*(\frac{3*65.45}{4pi})^(2/3) = -6.283\\V_{i+1} = 65.45-6.283*0.25 = 63.88$

i = 2, ti = 0.5, Vi = 63.88

$V'_{i} = -k *4pi*(\frac{3*63.88}{4pi})^(2/3) = -6.182\\V_{i+1} = 63.88-6.182*0.25 = 62.33$

i = 3, ti = 1, Vi = 62.33

$V'_{i} = -k *4pi*(\frac{3*62.33}{4pi})^(2/3) = -6.082\\V_{i+1} = 62.33-6.082*0.25 = 60.813$

We compute the next iterations in MATLAB (see attachment)

Volume @ t = 10 is = 20.2969

The droplet radius at t=10 mins

$r(10) = (\frac{3*20.2969}{4pi})^(2/3) = 1.692 mm\\$

The average change of droplet radius with time is:

Δr/Δt = $\frac{r(10) - r(0)}{10-0} = \frac{1.692 - 2.5}{10} = -0.0808 mm/min$

The value of the evaporation rate is close the value of k = 0.08 mm/min

Hence, the results are accurate and consistent!

5 0

3 years ago

Calculate the average speed (tin km/hr) of charlie who run to the store 4 km away in 30 minutes

Galina-37 [17]

ANSWER

$8km/hr$

EXPLANATION

The average speed

$= \frac{distance}{time}$

Distance=4km

Time=30mins

Convert time to hours.

Time =0.5hrs

The average speed

$= \frac{4}{0.5}$

$= 8km/hr$

8 0

3 years ago