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3 years ago

2x-4y=8 and 7x-3y=61 solve by substitution

2 answers:

8 0

2x= 8+4y. (Add 4y and move it to the other side)

X=8+4y/2 so X= 4+2y. (Divide both sides by 2)

Sub in 4+2y for x so 7(4+2y)-3y=61

Distribute: 28+ 14y -3y=61
Subtract 28 and add the y’s. 11y=33
Y=3

Ratling [72]3 years ago

4 0

Answer:

(10, 3)

Step-by-step explanation:

Solve by Substitution

2x − 4y = 8 and 7x − 3y = 61

Solve for x in the first equation.

x = 4 + 2y 7x − 3y = 61

Replace all occurrences of x with 4 + 2y in each e quation.

Replace all occurrences of x in 7x − 3y = 61 with 4 + 2y. 7 (4 + 2y) − 3y = 61

x = 4 + 2y

Simplify 7 (4 + 2y) − 3y.

28 + 11y = 61

x = 4 + 2y

Solve for y in the first equation.

Move all terms not containing y to the right side of the equation.

11y = 33

x = 4 + 2y

Divide each term by 11 and simplify.

y = 3

x = 4 + 2y

Replace all occurrences of y with 3 in each equation.

Replace all occurrences of y in x = 4 + 2y with 3. x = 4 + 2 (3)

y = 3

Simplify 4 + 2 (3).

x = 10

y = 3

The solution to the system is the complete set of ordered pairs that are valid solutions.

(10, 3)

The result can be shown in multiple forms.

Point Form:

(10, 3)

Equation Form:

x = 10, y = 3

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Answer:

2π feet

Step-by-step explanation:

9:30 PM to 1:30 AM is 4 hours.

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A full 12 hours would be 2×r×π=2×3×π

1/3 of that is:

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3 0

3 years ago

Need help with inscribed angles If M

Bogdan [553]

Given:

In the given circle O, BC is diameter, OA is radius, DC is a chord parallel to chord BA and $m\angle BCD=30^\circ$ .

To find:

The $m\angle AOB$ .

Solution:

If a transversal line intersect two parallel lines, then the alternate interior angles are congruent.

We have, DC is parallel to BA and BC is the transversal line.

$\angle OBA\cong \angle BCD$ [Alternate interior angles]

$m\angle OBA=m\angle BCD$

$m\angle OBA=30^\circ$

In triangle AOB, OA and OB are radii of the circle O. It means OA=OB and triangle AOB is an isosceles triangle.

We know that base angles of an isosceles triangle are congruent.

$\angle OAB\cong \angle OBA$ [Base angles of an isosceles triangle]

$m\angle OAB=m\angle OBA$

$m\angle OAB=30^\circ$

In triangle AOB,

$m\angle OAB+m\angle OBA+m\angle AOB=180^\circ$

$30^\circ+30^\circ+m\angle AOB=180^\circ$

$60^\circ+m\angle AOB=180^\circ$

$m\angle AOB=180^\circ-60^\circ$

$m\angle AOB=120^\circ$

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6 0

2 years ago

Un curso esta formado por 12 mujeres y 15 hombres ¿cual es la probabilidad de que las personas en el cargo de presidente y visep

riadik2000 [5.3K]

Answer:

P = (12/27)*(11/26) = 0.188

Step-by-step explanation:

Si todos tienen la misma probabilidad de estar en cargo de presidente y vicepresidente, entonces:

La probabilidad de que una mujer sea presidente, es igual al cociente entre el número de mujeres y el número total de personas:

p = 12/27

Ahora queremos calcular lo mismo para vicepresidente, y el cálculo es exactamente igual, con la diferencia de que en esta situación, ya hay una mujer en el cargo de presidente, entonces hay 11 mujeres que pueden ser vicepresidente, y un total de: (11 + 15 = 26 personas)

Entonces ahora la probabilidad es:

q = 11/26

La probabilidad conjunta (es decir, la probabilidad de que ambos eventos pasen) es igual al producto de las probabilidades individuales.

P = (12/27)*(11/26) = 0.188

4 0

2 years ago

If m<mkl=83. m<jkl=127. and m<jkm=(9x-10). find the value of x

kirza4 [7]

MKL = 83, JKL = 127, JKM = 9x - 10 given

JKL + MKL = JKM angle addition postulate

127 + 83 = 9x - 10 substitution

210 = 9x - 10 simplify (add like terms)

220 = 9x addition property of equality

$\frac{220}{9}$ = x

3 0

3 years ago

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KiRa [710]

Answer:

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3 years ago