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julsineya [31]
3 years ago
5

Solve the following expression and state the operation used in each step. Show all of your work 5+9-3^2+2​

Mathematics
1 answer:
Pepsi [2]3 years ago
8 0
Answer: 7


5+9-3^2+2
= 14-3^2+2
= 14-9+2
= 5+2
= 7
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Pls, help me with this math question
olya-2409 [2.1K]

Answer:

Perimeter \sqrt{26}+\sqrt{5}+\sqrt{10}+\sqrt{17} units.  Area 12 square units.

Step-by-step explanation:

Perimeter: total distance around the figure.

Distance Formula:  the distance between points \left(x_1,y_1\right) \text{ and } \left(x_2,y_2\right) is

d=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}

AB=\sqrt{(6-1)^2+(2-1)^2}=\sqrt{25+1}=\sqrt{26}

BC=\sqrt{(5-6)^2+(4-2)^2}=\sqrt{1+4}=\sqrt{5}CD=\sqrt{(2-5)^2+(5-4)^2}=\sqrt{9+1}=\sqrt{10}DA=\sqrt{(1-2)^2+(1-5)^2}=\sqrt{1+16}=\sqrt{17}

The perimeter is the sum of all those segment lengths.

One way to find the area of the figure is to surround it with a rectangle, insert some lines so that the areas you do not want can be found and subtracted from the rectangle's area.  (See attached image.)

The area of the large rectangle around the figure is 5 x 4 = 20 square units.

The triangles have areas 1/2 (base) (height):

A. (1/2)(1)(4) = 2 square units

B. (1/2)(3)(1) = 1.5 square units

D. (1/2)(1)(2) = 1 square unit

E. (1/2)(5)(1) = 2.5 square units

Square C.  (1)(1) = 1 square unit

Total of all the area you don't want to include:

2 + 1.5 + 1 + 2.5 + 1 = 8 square units

Subtract 8 from the surrounding rectangle's area of 20, and you get the area of the figure is 20 - 8 = 12 square units.

4 0
3 years ago
CAN SOMEONE PLEASE PLEASE PLEASE HELP ME, YOU’LL GET FREE EASY POINTS IF YOU GIVE ME THE RIGHT ANSWER !!
bekas [8.4K]

Answer:

  1. reflection across BC
  2. the image of a vertex will coincide with its corresponding vertex
  3. SSS: AB≅GB, AC≅GC, BC≅BC.

Step-by-step explanation:

We want to identify a rigid transformation that maps congruent triangles to one-another, to explain the coincidence of corresponding parts, and to identify the theorems that show congruence.

__

<h3>1.</h3>

Triangles GBC and ABC share side BC. Whatever rigid transformation we use will leave segment BC invariant. Translation and rotation do not do that. The only possible transformation that will leave BC invariant is <em>reflection across line BC</em>.

__

<h3>2.</h3>

In part 3, we show ∆GBC ≅ ∆ABC. That means vertices A and G are corresponding vertices. When we map the congruent figures onto each other, <em>corresponding parts are coincident</em>. That is, vertex G' (the image of vertex G) will coincide with vertex A.

__

<h3>3.</h3>

The markings on the figure show the corresponding parts to be ...

  • side AB and side GB
  • side AC and side GC
  • angle ABC and angle GBC
  • angle BAC and angle BGC

And the reflexive property of congruence tells us BC corresponds to itself:

  • side BC and side BC

There are four available congruence theorems applicable to triangles that are not right triangles

  • SSS -- three pairs of corresponding sides
  • SAS -- two corresponding sides and the angle between
  • ASA -- two corresponding angles and the side between
  • AAS -- two corresponding angles and the side not between

We don't know which of these are in your notes, but we do know that all of them can be used. AAS can be used with two different sides. SAS can be used with two different angles.

SSS

  Corresponding sides are listed above. Here, we list them again:

  AB and GB; AC and GC; BC and BC

SAS

  One use is with AB, BC, and angle ABC corresponding to GB, BC, and angle GBC.

  Another use is with BA, AC, and angle BAC corresponding to BG, GC, and angle BGC.

ASA

  Angles CAB and CBA, side AB corresponding to angles CGB and CBG, side GB.

AAS

  One use is with angles CBA and CAB, side CB corresponding to angles CBG and CGB, side CB.

  Another use is with angles CBA and CAB, side CA corresponding to angles CBG and CGB, side CG.

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<span>The initial value is the value of the y-coordinate when the x-coordinate is equal to 0. That is the same to say the value at which the graph intercepts the y-axis. So, you justt have to pick the graph whose line intercepts the y-axis at 1/2. </span>
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Answer:

2

Step-by-step explanation:

I hope this helps!

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