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r-ruslan [8.4K]
3 years ago
11

Practice test

Mathematics
2 answers:
Elan Coil [88]3 years ago
7 0

Answer:

48

Step-by-step explanation:

mariarad [96]3 years ago
3 0

Answer:

it's a rectangle so the area is 6×8=48

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Use order of operation to solve 4[2.5+3(20+7)]
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334 is the answer to the question

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11k - 8 = 11 + 8 ( k + 9 )
Alexus [3.1K]

Answer:

k = \frac{91}{3}

Step-by-step explanation:

Expand.

11k - 8 = 11 + 8k + 72

Simplify 11 + 8k + 72 to 8k + 83.

11k - 8 = 8k + 83

Add 8 to both sides.

11k + 8k + 83 + 8

Simplify 8k + 83 + 8 to 8k + 91.

11k = 8k + 91

Subtract 8k from both sides.

11k - 8k = 91

Simplify 11k - 8k to 3k.

3k = 91

Divide both sides by 3.

k = \frac{91}{3}

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3 years ago
Solve -7 - | x- 6| = - 12
andrezito [222]

Answer:

x=54/ 7 , 30/7

Step-by-step explanation:

x=54/ 7 , 30/7

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This is a bit confusing for me. I understand how the whole X axis and Y axis works, and various graphs, but for some reason the
Dafna1 [17]
Your coordinates are (x,y), here you just need to figure what 'x' is given that 'y' is equal to -3. So, just plug in -3 for 'y' in the equation and then solve for 'x', like so: 4x + 5(-3) = -7 ... then just use algebra to solve for what 'x' has to be. Hope this helps!
7 0
3 years ago
Una torre de 28.2 m de altura esta situada a la orilla de un rio, desde lo alto del edificio el ángulo de depresión a la orilla
Svet_ta [14]

Answer:

El ancho del río es 59.9 metros.

Step-by-step explanation:

El ancho del río lo podemos calcular con la siguiente relación trigonométrica asumiendo que la torre forma un triángulo rectángulo con el río:

tan(\theta) = \frac{CO}{CA}

En donde:

CA: es el cateto adyacente = Altura de la torre = 28.2 m

CO: es el cateto opuesto = ancho del río =?

θ: es el ángulo adyacente a CA

Dado que el ángulo de depresión (25.2°) está ubicado fuera de la parte superior de la hipotenusa del triángulo que forma la torre con la orilla opuesta del río, debemos calcular el ángulo interno (θ) como sigue:

\theta = (90 - 25.2)^{\circ} = 64.8 ^{\circ}

Ahora, el ancho del río es:

CO = tan(\alpha)*CA = tan(64.8)*28.2 = 59.9 m

Por lo tanto, el ancho del río es 59.9 metros.

Espero que te sea de utilidad!                  

4 0
3 years ago
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