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Delvig [45]
3 years ago
9

Enzo says that he can draw an enlarged rectangle that is 16 centimeters by 13 centimeters. Which explains whether Enzo is correc

t?
Mathematics
2 answers:
Zarrin [17]3 years ago
6 0
He cannot make a rectangle with those measurements because the base of the rectangle need to be a least more that 1cm bigger than the height other wise its just a square  
nikklg [1K]3 years ago
5 0
The scale drawing is a representative drawing that is supposed to depict the ratio between the sides or dimensions of the actual figure. For example, given that the scale factor is 0.5, the 16 centimeters by 13 centimeters actual drawing should be drawn as 8 cm by 6.5 centimeters. 
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How many solutions does the equation 2x-6=6+2x
Alex_Xolod [135]

Answer:

2 solutions

Step-by-step explanation:

Since the equations are not equal, they provide two solutions. Negative numbers provide no solutions. If they are equal to zero, they have one solution.


5 0
4 years ago
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What addds to 14 and multiplies to 8
Grace [21]
5.4142 and <span>2.5858 are the numbers you need</span>
7 0
3 years ago
This is converting to polar form, I need help and the answer too.
Mila [183]

Answer:

2 cis (7/6 pi)

Step-by-step explanation:

r = sqrt( a^2 + b^2)

r = sqrt (-sqrt(3) )^2 + (-1)^2)

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    = sqrt(4)

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theta  = arctan (b/a)

theta = arctan (-1/-sqrt(3))

theta = 30

but this is in the third quardrant  -a and -b

so add 180

theta = 210 degrees

convert this to radians

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r cis (theta)

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7 0
3 years ago
-<br>nun<br>THE SUM OF THREE TIMES A<br>Numbers and 11​
Vikentia [17]

Answer:

3x + 11

Step-by-step explanation:

Remember BPEMDAS.

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"The sun of 3x and 11" is saying add our 3x and 11, so: 3x + 11

8 0
3 years ago
What are the solutions to this equation?
ziro4ka [17]

 

\displaystyle\\(x-4)^2=49\\\\(x-4)^2-49=0\\\\(x-4)^2-7^2=0\\\\(x-4-7)(x-4+7)=0\\\\(x-11)(x+3)=0\\\\x-11 = 0~~~\text{or}~~~x+3=0\\\\x-11=0~~~\implies~~~\boxed{x_1=11}\\\\x+3=0~~~\implies~~~\boxed{x_2=-3}\\\\\boxed{\bf The~solutions~are:~~x =11~~\text{or}~~x =-3}



8 0
3 years ago
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