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CaHeK987 [17]
3 years ago
8

Mrs. Valdez wanted to determine the number of people, p, she could safely take in her car to the fun run. She determined that p

< 5. Which statement best describes a possible number of people she can take in her car?
Mrs. Valdez can take –2 people because –2 < 5.

Mrs. Valdez can take 2 people because 2 < 5.

Mrs. Valdez can take 3.5 people because 3.5 < 5. NOT THIS

Mrs. Valdez can take 5 people because 5 < 5.


its not c


What is the first step in solving the inequality < –1?


Multiply both sides by 6.

Add 2 to both sides. NOT THIS ONE

Change the direction of the inequality.

Change the inequality to ≤.
Mathematics
1 answer:
dolphi86 [110]3 years ago
5 0

Answer:

for the first question b,


it doesn't make sense for choice a to have -2 people in the car. You can't have a negative amount of people. For choice choice c, you can't have half a person, and for choice d, Ms. Valdez can only have under 5 people, not five people exactly.


for the second question I need to see the whole inequality, generally there is a variable, and all i see is <-1, which is invalid.


hope this helps, if it does like it pls!!


Step-by-step explanation:


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Answer:

\triangle EDF is isosceles.

Step-by-step explanation:

Please have a look at the attached figure.

We are <u>given</u> the following things:

\angle EDF = y

\text{External }\angle DFG = 90 +\dfrac{y}{2}

Let us try to find out \angle E and \angle DFE. After that we will compare them.

<u>Finding </u>\angle DFE<u>:</u>

Side EG is a straight line so \angle GFE = 180

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\angle GFE = 180 = \angle DFE  + \angle DFG\\\Rightarrow 180 = \angle DFE + (90+\dfrac{y}{2})\\\Rightarrow \angle DFE = 180 - 90 - \dfrac{y}{2}\\\Rightarrow \angle DFE = 90 - \dfrac{y}{2} ....... (1)

<u>Finding </u>\angle E<u>:</u>

<u>Property of external angle:</u> External angle in a triangle is equal to the sum of two opposite internal angles of a triangle.

i.e. external \angle DFG = \angle E + \angle EDF

\Rightarrow 90+\dfrac{y}{2} = \angle E + y\\\Rightarrow \angle E = 90+\dfrac{y}{2}  -y\\\Rightarrow \angle E = 90-\dfrac{y}{2} ....... (2)

Comparing equations (1) and (2):

It can be clearly seen that:

\angle DFE = \angle E =90-\dfrac{y}{2}

The two angles of \triangle EDF are equal hence \triangle EDF is isosceles.

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Answer:

Step-by-step explanation:

Top Problem:

Reason:

1. Given

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