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Elden [556K]
2 years ago
13

Select the correct answer.

Mathematics
1 answer:
Allushta [10]2 years ago
5 0

Answer:

Step-by-step explanation:

Like terms have same variable with same exponent. Combine the like terms.

x + x = 2x  {Just add the coefficient}

x + y + x +y + 30 + 5 = <u>x +x</u>  +  <u>y + y</u>   + <u>30 + 5</u>

                                 = 2x + 2y + 35

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Determine if the conditional and its converse are true. If they are both true, select which biconditional correctly represents t
Ugo [173]
The conditional,  <span>If four points are non-coplanar, then they are non-collinear, </span>is true:

This is, coplanarity is a necessary condition to be collinear.

The converse, <span>If four points are non-collinear, then they are non-coplanar, is false.

A counterexample that disproves this statement is the 4 vertices of a paralelogram, of course they are in a same plane and are not collinear.
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3 0
3 years ago
Help me plz &lt;3 I need help
quester [9]
180-47=133 bc straight angles measure 180
6 0
3 years ago
LAST QUESTION I NEED! Should be easy (I don’t think the answer is 4 might be wrong if given an explanation)
mina [271]

Answer:

2

Step-by-step explanation:

because 2 square is 4 then that was needed for the other squares but this one is doubled

5 0
3 years ago
Q‒1. [5×4 marks] a) How many three-digit numbers can be formed from the digits 0, 1, 2, 3, 4, 5, and 6? (150) b) How many three-
amid [387]

Answer:

a) 294

b) 180

c) 75

d) 174

e) 105

Step-by-step explanation:

I assume that for each problem, the first digit can't be 0.

a) There are 6 digits that can be first, 7 digits that can be second, and 7 digits that can be third.

6×7×7 = 294

b) This time, no digit can be used twice, so there are 6 digits that can be first, 6 digits that can be second, and 5 digits that can be third.

6×6×5 = 180

c) Again, each digit can only be used once, but this time, the last digit must be odd.

If only the last digit is odd, there are 3×3×3 = 27 possible numbers.

If the first and last digits are odd, there are 3×4×2 = 24 possible numbers.

If the second and last digits are odd, there are 3×3×2 = 18 possible numbers.

If all three digits are odd, there are 3×2×1 = 6 possible numbers.

The total is 27 + 24 + 18 + 6 = 75.

d) If the first digit is 3, and the second digit is 3, there are 1×1×6 = 6 possible numbers.

If the first digit is 3, and the second digit is greater than 3, there are 1×3×7 = 21 possible numbers.

If the first digit is greater than 3, there are 3×7×7 = 147 numbers.

The total is 6 + 21 + 147 = 174.

e) If the first digit is 3, and the second digit is greater than 3, then there are 1×3×5 = 15 possible numbers.

If the second digit is greater than 3, there are 3×6×5 = 90 possible numbers.

The total is 15 + 90 = 105.

6 0
3 years ago
In the following equations, what would you multiply the second equation to
Leni [432]

Answer:-3

Step-by-step explanation:

6 0
2 years ago
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