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Oduvanchick [21]
3 years ago
15

.please help me with this I will crown you brainlest

Mathematics
2 answers:
Brut [27]3 years ago
8 0
I think it is either A or C
but more sure about A
Marizza181 [45]3 years ago
4 0
You can prove it by the SAS postulate which should be the first option
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Brett and Max are diving. Brett is 34 feet below the surface and Max is 25 feet below the surface. How many feet above Brett isM
saveliy_v [14]

Answer:

Max is 9 ft above Brett

Step-by-step explanation:

Brett is 34 feet below the surface  = -34

Max is 25 feet below the surface = -25

Subtract the heights

-25 - -34

-25 +34

9

Max is 9 ft above Brett

6 0
3 years ago
Read 2 more answers
The length of a rectangle is 5 metres less than twice the breadth. If the perimeter is 50 meters,find the length and breadth
MAXImum [283]
<h3><u>S</u><u> </u><u>O</u><u> </u><u>L</u><u> </u><u>U</u><u> </u><u>T</u><u> </u><u>I</u><u> </u><u>O</u><u> </u><u>N</u><u> </u><u>:</u></h3>

As per the given question, it is stated that the length of a rectangle is 5 m less than twice the breadth.

Assumption : Let us assume the length as "l" and width as "b". So,

\twoheadrightarrow \quad\sf{ Length =2(Width)-5}

\twoheadrightarrow \quad\sf{ \ell=(2b-5) \; m}

Also, we are given that the perimeter of the rectangle is 50 m. Basically, we need to apply here the formula of perimeter of rectangle which will act as a linear equation here.

\\ \twoheadrightarrow \quad\sf{ Perimeter_{(Rectangle)} = 2(\ell +b) } \\

  • <em>l</em> denotes length
  • <em>b</em> denotes breadth

\\ \twoheadrightarrow \quad\sf{50= 2(2b-5+b)} \\

\\ \twoheadrightarrow \quad\sf{50= 2(3b-5)} \\

\\ \twoheadrightarrow \quad\sf{50= 6b - 10} \\

\\ \twoheadrightarrow \quad\sf{50+10= 6b} \\

\\ \twoheadrightarrow \quad\sf{60= 6b} \\

\\ \twoheadrightarrow \quad\sf{\cancel{\dfrac{60}{6}}=b} \\

\\ \twoheadrightarrow \quad\underline{\bf{10\; m = Width }} \\

Now, finding the length. According to the question,

\twoheadrightarrow \quad\sf{ \ell=(2b-5) \; m}

\twoheadrightarrow \quad\sf{ \ell=2(10)-5\; m}

\twoheadrightarrow \quad\sf{ \ell=20-5\; m}

\\ \twoheadrightarrow \quad\underline{\bf{15\; m = Length }} \\

<u>Therefore</u><u>,</u><u> </u><u>length</u><u> </u><u>and</u><u> </u><u>breadth</u><u> </u><u>of</u><u> </u><u>the</u><u> </u><u>r</u><u>ectangle</u><u> </u><u>is</u><u> </u><u>1</u><u>5</u><u> </u><u>m</u><u> </u><u>and</u><u> </u><u>10</u><u> </u><u>m</u><u>.</u><u> </u>

7 0
3 years ago
lee packed 2 sweatshirts and 6 t shirts in her suitcase. Which ration doe not represent the number of sweatshirts to the number
aleksley [76]

Answer: c

Step-by-step explanation:

The ratio is not 3:1

7 0
3 years ago
Some help would be great, also giving brainliest.
Harrizon [31]
The area of the shaded region would be 64m²
8 0
3 years ago
Determine the value of x in the equation x+5x=5x+7
12345 [234]

\text{Hey there!}

\mathsf{Determine\ the\ value\ of\ x\ in\ the\ equation\ x+5x=5x+7.}

\mathsf{SIMPLIFY\ both\ sides\ of\ your\ equation}

\mathsf{x+5x=5x+7}

\mathsf{COMBINE\ your\ like\ terms}

\mathsf{(x+5x)=(5x+7)}

\mathsf{The\ RIGHT\ side\ DOES\  NOT\ have\ any\ like\ terms\ so\ we\ can}\\\mathsf{can\ leave\ alone}

\mathsf{x+5x= 6x}

\mathsf{Your\ new\ equation: 6x=5x+7}

\mathsf{SUBTRACT\ by\ 5x\ on\ your\ sides}

\mathsf{6x-5x}\\\mathsf{=}\\\mathsf{7-5x}

\mathsf{SOLVE\ your\ equation\ above\uparrow}

\mathsf{If\ you\ SOLVED\ correctly\ you\ SHOULD\ have\ the\ VALUE\ of\ x}

\boxed{\boxed{\mathsf{Thus\ your\ answer\ is\ x=7}}}\checkmark

\text{Good luck on your assignment and enjoy your day!}

~\dfrac{\frak{LoveYourselfFirst}}{:)}

6 0
3 years ago
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