Can you please help me?

(1.2)Consider the diagram below. Identify a point.

Fynjy0 [20]

Option B is correct. Point F is the only point according to the option given

A point is defined as an exact location with no size, length, or even width.

From the diagram shown, we have points, plane, and line segments.

The only plane that we have is plane R making option C incorrect.

A line is defined as a straight figure connected by two points. From the diagram, the line segments that we have are AN, BN, and AB falsifying options A and D.

The points are C, D, E, and F. Hence the correct option is B. Point F is the only point according to the option given

Learn more here: brainly.com/question/13099718

6 0

3 years ago

Question 10

kicyunya [14]

Answer:

47.5 and 2853.56 m

Step-by-step explanation:

Graph the equation on desmos and use the graph to find the maxima and the roots

6 0

2 years ago

Read 2 more answers

John enlarged one side of his square flower bed by 8 ft to make it into a rectangle. The area of the new flower bed is 3 times t

kobusy [5.1K]

Answer:

4

Step-by-step explanation:

4*4=16

(4+8)4=48

16*3=16!

5 0

2 years ago

Read 2 more answers

Colin has a pad with x pieces of paper on it. For his first class, he wrote on 5 fewer than half of the pieces of paper in the p

PIT_PIT [208]

Answer:

Colin has 8 sheets left for his third class.

Step-by-step explanation:

Given that:

Total Number of pieces of papers = $x$

Number of pieces of papers used for 1st class = 5 fewer than half of the pieces in the pad

Writing the equation:

$\text{Number of pieces of papers used for 1st class =} \dfrac{x}{2} -5 ...... (1)$

Also, Given that number of pieces of papers used for the 2nd class are 2 more than that of papers used in the 1st class.

$\text{Number of pieces of papers used for 2nd class =} \dfrac{x}{2} -5+2 = \dfrac{x}2 -3 ...... (2)$

Now, number of pieces of papers left for the third class = Total number of pieces of papers in the pad - Number of pieces of papers used in the first class - Number of pieces of papers used in the first class

$\text{number of pieces of papers left for the third class = }x-(\dfrac{x}{2}-5)-(\dfrac{x}{2}-3)\\\Rightarrow x-\dfrac{x}2-\dfrac{x}2+5+3\\\Rightarrow x-x+5+3\\\Rightarrow 8$

So, the answer is:

Colin has 8 sheets left for his third class.

5 0

3 years ago

Solve 4 over x minus 4 equals the quantity of x over x minus 4, minus four thirds for x and determine if the solution is extrane

Vinil7 [7]

I'm thinking this is what the problem looks like: $\frac{4}{x-4}= \frac{x}{x-4}- \frac{4}{3}$ . The first thing to do is to move the $\frac{x}{x-4}$ over to the other side because it has a common denominator with the other side. Doing that and at the same time combining them over their common denominator looks like this: $\frac{4-x}{x-4}= -\frac{4}{3}$ . The best way to solve for x now is to cross-multiply to get 3(4-x)=-4(x-4). Distributing through the parenthesis is 12 - 3x = -4x + 16. Solving for x gives us x = 4. Of course when we sub a 4 back in for x we get real problems, don't we? Dividing by zero breaks every rule in math that there ever was! So, yes, the solution is extraneous.

6 0

3 years ago