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Eduardwww [97]
3 years ago
14

Need help with math pls help me will mark brainiest.

Mathematics
1 answer:
Greeley [361]3 years ago
8 0

Answer:

1. B.  Slope-intercept form: y = \frac{1}{5}x - 2

2. A. \frac{1}{2}

Step-by-step explanation:

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Can someone help me with this question:
ki77a [65]
A.)The equation is 24 + 8=32. Twice of the amount of 24 is 6 times the amount of 8. 24 x 2= 48. 48/6=8.
B.) X +Y=32 and 2x=6y

4 0
3 years ago
Solve this equation<br> 3x - 5 = -26​
Dmitry_Shevchenko [17]

3x - 5 = -26​

<u>step 1:</u>

subtract +5 on each side because you have to do the inverse operation

<u>step 2:</u>

3x= -21

<u>step 3:</u>

now divide 3 on each side

the answer is -7

<u />

Hope this helped!

<u />

<em>TheOneAndOnlyLara~</em>

5 0
1 year ago
Read 2 more answers
C) Parbati buys a mobile for Rs 6,300 and sells it to Laxmi at 15% profit. How much
Leni [432]

Answer:

Rs. 7245

Step-by-step explanation:

Given parameters:

Cost price  = Rs. 6300

Percentage profit  = 15%

Unknown:

Selling price = ?

Solution:

If profit is made on a trade, the selling price is higher than the cost price.

 Profit  = Selling price  - Cost price

To find the selling price simply;

  Selling price  =( 1 + \frac{15}{100})  x cost price

  Selling price  = 1.15 x 6300  = Rs. 7245

6 0
3 years ago
Solve. Justify your responses. a Given:a║b and c║d, m∠ 4=35° Find: m∠1, m∠2, and m∠3
STALIN [3.7K]

Answer:

<em>m∠1 =  90°</em>

<em>m∠2 = 35°</em>

<em>m∠3 = 55°</em>

Step-by-step explanation:

Find the diagram attached

From the diagram:

m

Given that <4 = 35°

m

The sum of the angle on a straight line at point A is 180°. Hence:

90+m

The sum of angle in the triangle ABC is 180°, hence;

m

7 0
2 years ago
1) UN MOVIL A SE MUEVE DESDE UN PUNTO CON VELOCIDAD CONSTANTE DE 20m/s EN EL MISMO INSTANTE A UNA DISTANCIA DE 1200m, OTRO MOVIL
alisha [4.7K]

Answer:

El móvil B necesita 60 segundos para alcanzar al móvil A y le alcanza una distancia de 2400 metros con respecto al punto de referencia.

Step-by-step explanation:

Supóngase que cada movil viaja en el mismo plano y que el móvil B se localiza inicialmente en la posición x = 0\,m, mientras que el móvil A se encuentra en la posición x = 1200\,m. Ambos móviles viajan a rapidez constante. Si el móvil B alcanza al móvil A después de cierto tiempo, el sistema de ecuaciones cinemáticas es el siguiente:

Móvil A

x_{A} = 1200\,m+\left(20\,\frac{m}{s} \right)\cdot t

Móvil B

x_{B} = \left(40\,\frac{m}{s} \right)\cdot t

Donde:

x_{A}, x_{B} - Posiciones finales de cada móvil, medidas en metros.

t - Tiempo, medido en segundos.

Si x_{A} = x_{B}, el tiempo requerido por el móvil B para alcanzar al móvil A es:

1200\,m+\left(20\,\frac{m}{s} \right)\cdot t = \left(40\,\frac{m}{s} \right)t

1200\,m = \left(20\,\frac{m}{s} \right)\cdot t

t = \frac{1200\,m}{20\,\frac{m}{s} }

t = 60\,s

El móvil B necesita 60 segundos para alcanzar al móvil A.

Ahora, la distancia se obtiene por sustitución directa en cualquiera de las ecuaciones cinemáticas:

x_{B} = \left(40\,\frac{m}{s} \right)\cdot (60\,s)

x_{B} = 2400\,m

El móvil B alcanza al móvil A a una distancia de 2400 metros con respecto al punto de referencia.

3 0
3 years ago
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