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Lostsunrise [7]
3 years ago
6

PLEASE HELP ME WITH THESE TWO

Mathematics
1 answer:
Artist 52 [7]3 years ago
4 0

7.

\begin{array}{l|l}{\underline {Statement} &\underline{Reason}\\1.\ MA\ ||\ RY&1.\ \text{Given}\\2.\ \angle M \cong \angle R&2.\ \text{Alternate Interior Angles Theorem}\\3.\ \angle M \cong \angle R&3.\ \text{Given}\\4.\ AY \cong AY&4.\ \text{Reflexive Property}\\5.\ \triangle MAY \cong \triangle RYA&5.\ \text{Angle-Angle-Side Theorem}\\6.\ MY \cong RA&6.\ \text{CPCTC}\\\end{array}


9.

\begin{array}{l|l}{\underline {Statement} &\underline{Reason}\\1.\ \triangle JKM \text{is equilangular}&1.\ \text{Given}\\2.\ \angle J \cong \angle K \cong \angle M&2.\ \text{De-finition of Equiangular}\\3.\ IE \perp MK&3.\ \text{Given}\\4.\ \angle IEK\ \text{and}\ \angle IEM\ \text{are right angles}&4.\ \text{De-finition of Perpendicular}\\5.\ \angle IEK \cong \angle IEM&5.\ \text{Transitive Property}\\6.\ IE \cong IE&6.\ \text{Reflexive Property}\\\end{array}

\begin{array}{l|l}\\7.\ \angle EIK\ \text{and}\ \angle EIM\ \qquad \qquad \qquad \quad &7.\ \text{Triangle Sum Theorem}\\8.\ \angle EIK \cong \angle EIM&8.\ \text{Transitive Property}\\9.\ \triangle IKE \cong \triangle IME&9.\ \text{Angle-Side-Angle Theorem}\end{array}


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So the answer will be
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Svetach [21]

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4 0
3 years ago
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Mary scored 28 points, which is twice as many points as Lashawna scored. Let L represent the points Lashawna scored.
Basile [38]

Answer: Lashawna scored 14 points

Step-by-step explanation:

Let M = Mary's score

L = Lashawna score

M = 28 points

And M is twice as much as L score

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8 0
4 years ago
2. Mr. Lavi determined that the total cost of his son's birthday party, c, could be represented by
Tresset [83]

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4 0
3 years ago
Need help with 21-23 you suppose to simplify the expression
lora16 [44]

To solve these problems, we must remember the distributive property. This property states that a coefficient being multiplied by a polynomial in parentheses is equal to the sum of the coefficient times each of the separate terms. Using this knowledge, let's begin with number 21:

-(4x + 17) + 3(7-x)

To begin, we should distribute the negative sign through the first set of parentheses and the coefficient of positive 3 through the second set of parentheses.

-4x - 17 + 21 - 3x

Next, we must combine like terms, or add/subtract the constants terms and the variable terms in order to create a more concise expression.

-7x + 4 (your answer)

Now, we can move on to question 22 and solve it in a similar manner:

7(2n-8) - 4(12 - 8n)

Again, we will distribute the coefficients through the parentheses. However, keep in mind that the coefficient in front of the second set of parentheses is actually a NEGATIVE 4, so we must distribute the negative as well.

14n - 56 - 48 + 32n

Next, we will combine like terms (add the n terms together and subtract the constant terms).

46n - 104

Now, we can solve problem 23:

8 + 2(5f - 3)

We will again distribute through the parentheses:

8 + 10f - 6

Combine like terms after that:

10f + 2

Therefore, your answers for the three problems are as follows:

21) -7x + 4

22) 46n - 104

23) 10f + 2

Hope this helps!

7 0
3 years ago
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