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3 years ago

1/4(3c+5)-1/2(2c+3)=1/2 Please show some work so I can fiqure it out.

2 answers:

4 0

So,

1/4(3c + 5) + -1/2(2c + 3) = 1/2

First, we distribute to get rid of the parentheses. (PEMDAS)
3/4c + 1 1/4 - c - 1 1/2 = 1/2

Now, we collect like terms.
-1/4c - 1/4 = 1/2

Add 1/4 to both sides.
-1/4c = 3/4

Now, multiply by the reciprocal (-4/1).
c = -3

Check your answer.

1/4(3[-3] + 5) + -1/2(2[-3] + 3) = 1/2

1/4(-9 + 5) + -1/2(-6 + 3) = 1/2

-2 1/4 + 1 1/4 + 3 - 1 1/2 = 1/2

-1 + 1 1/12 = 1/2

1/2 = 1/2

Novosadov [1.4K]3 years ago

4 0

1/4(3c+5)-1/2(2c+3)=1/23/4c+1u1/4-1c-1u1/2=1/2
3/4c-1c=-1/4c
1u1/4-1u1/2=-1/4
-1/4c-1/4=1/2
+1/4 +1/4
-1/4c=3/4
/-1/4 /-1/4
answer: -3

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Step-by-step explanation:

12a. Addition

.. 3y⁵ + 2y³ + y + 3

+ (4y⁵ + 5y³– 7y – 8)

————————————

= 7y⁵ + 7y³ – 6y – 5

12b. Subtraction

.. 3y⁵ + 2y³ + y + 3

– (4y⁵ + 5y³– 7y – 8)

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

= – y⁵ – 3y³ + 8y + 11

13a. Addition

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+ (3x⁴ + 5x³ +...................+ 7)

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

= 5x⁴ + x³ – 6x² + 8x + 17

13b. Subtraction

.. 2x⁴ – 4x³ – 6x² + 8x + 10

– (3x⁴ + 5x³ +...................+ 7)

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

= – x⁴ – 9x³ – 6x² + 8x + 3

14a. Addition

...... b⁴ + 4b³ – 6b² + 4b – 1

+ (– 3b⁴ – 3b³ – 5b² – 4b)

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

= – 2b⁴ + b³ – 11b² – 1

14b. Subtraction

...... b⁴ + 4b³ – 6b² + 4b – 1

– (– 3b⁴ – 3b³ – 5b² – 4b)

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

= 4b⁴ + 7b³ – b² + 8b – 1

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... m⁵ + m³ – 5

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¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

= m⁵ + m⁴ + m³ + m²

15b. Subtraction

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¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

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¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

= 4x⁴ + 12x³ + 18x² + 16x + 4

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¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

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Answer:

Step-by-step explanation:

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we get common ratio =-4

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Explicit Formula = $-2*(-4)^{n-1}$

2) Comparing the given recursive formula $a_{n}=a_{n-1}*-2$

with standard recursive formula $a_{n}=a_{n-1}*r$

we get common ratio =-2

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Explicit Formula = $-4*(-2)^{n-1}$

3)Comparing the given recursive formula $a_{n}=a_{n-1}*3$

with standard recursive formula $a_{n}=a_{n-1}*r$

we get common ratio =3

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4)Comparing the given recursive formula $a_{n}=a_{n-1}*-4$

with standard recursive formula $a_{n}=a_{n-1}*r$

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Explicit Formula = $3*(-4)^{n-1}$

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with standard recursive formula $a_{n}=a_{n-1}*r$

we get common ratio =-4

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Explicit Formula = $-4*(-4)^{n-1}$

6)Comparing the given recursive formula $a_{n}=a_{n-1}*-2$

with standard recursive formula $a_{n}=a_{n-1}*r$

we get common ratio =-2

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Explicit Formula = $3*(-2)^{n-1}$

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with standard recursive formula $a_{n}=a_{n-1}*r$

we get common ratio =-5

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Explicit Formula = $4*(-5)^{n-1}$

8)Comparing the given recursive formula $a_{n}=a_{n-1}*-5$

with standard recursive formula $a_{n}=a_{n-1}*r$

we get common ratio =-5

8th term= $a_{1}*(r)^{n-1}=2*(-5)^{7} =-156250.$

Explicit Formula = $2*(-5)^{n-1}$

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3 years ago