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Serhud [2]
3 years ago
10

Cncncncncncncncncncn

Mathematics
1 answer:
anzhelika [568]3 years ago
6 0

Answer:

k

Step-by-step explanation:

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Multiply 5/12 and the multiplicative inverse of -1/4​
IgorC [24]

Answer:

The answer is -2

Step-by-step explanation:

Hope it helps :)

4 0
3 years ago
PLEASE HELP 13 INSTEAD OF FIVE POINTS!!!! I REALLY NEED HELP
devlian [24]
A is what the answer is
6 0
3 years ago
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12 of the 28 dogs at the shelter are purebred what percent of the dogs are pure breed?HELP!
harina [27]

Answer:

57.142%

Step-by-step explanation:

28-12=16

16/28*100=57.142%

3 0
3 years ago
what is the quotient (65y3 15y2 − 25y) ÷ 5y? a. 13y2 3y − 5 b. 13y3 3y2 − 5y c. 13y2 − 3y 5 d.13y2 − 3y − 5
Aneli [31]

Keywords:

<em>Division, quotient, polynomial, monomial </em>

For this case we must solve a division between a polynomial and a monomial and indicate which is the quotient.

By definition, if we have a division of the form: \frac {a} {b} = c, the quotient is given by "c".

We have the following polynomial:

65y ^ 3 + 15y ^ 2 - 25y that must be divided between monomy5y, then:

C (y) represents the quotient of the division:

C (y) = \frac {65y ^ 3 + 15y ^ 2 - 25y} {5y}

C (y) = \frac {65y ^ 3} {5y} + \frac {15y ^ 2} {5y} - \frac {25y} {5y}

C (y) = 13y ^ 2 + 3y-5

Thus, the quotient of the division between the polynomial and the monomial is given by:

C (y) = 13y ^ 2 + 3y-5

Answer:

The quotient is: C (y) = 13y ^ 2 + 3y-5

Option: A


4 0
3 years ago
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Help pls :’( ASAPP!!!<br> “Complete the proof”
nata0808 [166]

1) \overline{AB} \cong \overline{CD}, \overline{AD} \cong \overline{CB}, \overline{AX} \perp \overline{BD}, \overline{CY} \perp\overline{BD} (given)

2) \overline{BD} \cong \overline{BD} (reflexive property)

3) \triangle ABD \cong \triangle ACDB (SSS)

4) \angle ADB \cong \angle CBY (CPCTC)

5) \angle CYB and \angle AXD are right angles (perpendicular lines form right angles)

6) \triangle CYB and \triangle AXD are right triangles (a triangle with a right angle is a right triangle)

7) \triangle AXD \cong \triangle CYB (HA)

8) \overline{AX} \cong \overline{CY} (CPCTC)

6 0
2 years ago
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