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3 years ago

Jordan is making a paddleball court. The court consists of a wall outlined by 40 m of paint. What dimensions

Mathematics

2 answers:

ANTONII [103]3 years ago

6 0

Answer:

ffhhfhhfhhfhfhfhfhhfhhfhfhfhfhfhfhfhfhfhhfhfhfhfh

nydimaria [60]3 years ago

3 0

Answer:

lnfadvogbfvuvgayuvgalof

Step-by-step explanation:

<h3>asdbujvygadsoufyudgasoufgaedoy</h3>

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The amount $4.50 is 6% of what price

Vladimir79 [104]

Answer:

0.75

Step-by-step explanation:

4.50 is 6% of 75% then 0.75 is 75% in decimal form, so 0.75 is 6% of 4.50

Hope I helped :)

3 0

3 years ago

Read 2 more answers

Help with these algebra 2 questions

monitta

$Given\;:\sqrt{-18} - \sqrt{-32}$

we know that i² = -1

$\implies\sqrt{18}i - \sqrt{32}i\\$

$\implies 3\sqrt{2}i - 4\sqrt{2}i$

$\implies -i\sqrt{2}$

6 0

3 years ago

Read 2 more answers

Cual es la magnitud de la aceleración angular de una llanta de un automóvil que adquiere una velocidad angular de 350 rad/s en 2

drek231 [11]

Respuesta:

175 rad / s²

Explicación paso a paso:

Dado que :

Velocidad angular final = 350 rad / s

Tiempo empleado = 2 segundos

La aceleración angular, a =?

Recordar ; la aceleración angular es igual a la carga en velocidad angular con el tiempo

(velocidad final - Velocidad inicial) / tiempo empleado

Velocidad inicial = 0 rad / s

Aceleración angular = (350 - 0) / 2 = 350/2

= 175 rad / s²

3 0

2 years ago

What is 18.75 rounded two decimals places

Serggg [28]

The answer would be 18.8

7 0

3 years ago

If a cup of coffee has temperature 97Â°C in a room where the ambient air temperature is 25Â°C, then, according to Newton's Law o

FinnZ [79.3K]

Answer:

The average temperature is $T_{a} = 81.95^oC$

Step-by-step explanation:

From the question we are told that

The temperature of the coffee after time t is $T(t) = 25 + 72 e^{[-\frac{t}{45} ]}$

Now the average temperature during the first 22 minutes i.e fro $0 \to 22$ minutes is mathematically evaluated as

$T_{a} = \frac{1}{22-0} \int\limits^{22}_{0} {25 +72 e^{[-\frac{t}{45} ]}} \, dx$

$T_{a} = \frac{1}{22} [25 t + 72 [\frac{e^{[-\frac{t}{45} ]}}{-\frac{1}{45} } ] ] \left| 22} \atop {0}} \right.$

$T_{a} = \frac{1}{22} [25 t - 3240e^{[-\frac{t}{45} ]} ] \left | 45} \atop {{0}} \right.$

$T_{a} = \frac{1}{22} [25 (22) - 3240e^{[-\frac{22}{45} ]} - (- 3240e^{0} )]$

$T_{a} = \frac{1}{22} [550 - 1987.12 + 3240]$

$T_{a} = 81.95^oC$

8 0

3 years ago