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Advocard [28]
3 years ago
7

How would you prove quadrilateral WXYZ is a square? Is quadrilateral WXYZ a square explain why or why not.

Mathematics
1 answer:
WINSTONCH [101]3 years ago
4 0

Answer:

<em><u>If a quadrilateral has four congruent sides and four right angles, then it's a square</u></em>

If two consecutive sides of a rectangle are congruent, then it's a square

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Factor completely x2 − 49. (1 point) (x − 7)(x − 7) (x + 7)(x − 7) (x + 7)(x + 7) Prime
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Answer:

(x + 7)(x - 7)

8 0
3 years ago
A tree grows the same amount each year.one year the tree is 48 inches tall.When it is measured 3 years later it is 84 inches.How
slamgirl [31]
84-48=36
36/3=12
The tree grows 12 inches every year
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The interior angles formed by the sides of a quadrilateral have measures that sum to 360°.
stira [4]
The answer is 90 because 360/4 (since it is a quadrilateral) is 90 degrees
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2 years ago
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Wyatt solved the following equation:
ohaa [14]

Answer:

x = 2

Step-by-step explanation:

x + 1/2 (6x - 4) = 6

Expand the brackets.

x + 3x - 2 = 6

Add like terms.

4x - 2 = 6

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4x = 6 + 2

4x = 8

Divide 4 into both sides.

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5 0
3 years ago
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Can someone please answer these questions to help me understand? Please and thank you! Will mark as brainliest!!
Nadya [2.5K]

QUESTION 1  

If a function is continuous at x=a, then \lim_{x \to a}f(x)=f(a)  

Let us find the limit first,  

\lim_{x \to 4} \frac{x-4}{x+5}  

As x \rightarrow 4, x-4 \rightarrow 0,x+5 \rightarrow 9 and f(x) \rightarrow \frac{0}{9}=0  

\therefore \lim_{x \to 4} \frac{x-4}{x+5}=0  

Let us now find the functional value at x=4  

f(4)=\frac{4-4}{4+5} =\frac{0}{9}=0  

Since  

\lim_{x \to 4} f(x)=\frac{x-4}{x+5}=f(4), the function is continuous at a=4.  

QUESTION 2  

The correct answer is table 2. See attachment.


In this table the values of x approaches zero from both sides.


This can help us determine if the one sided limits are approaching the same value.

As we are getting closer and closer to zero from both sides, the function is approaching 2.


The values are also very close to zero unlike those in table 4.


The correct answer is B


QUESTION 3


We want to evaluate;


\lim_{x \to 1} \frac{x^3+5x^2+3x-9}{x-1}


using the properties of limits.


A direct evaluation gives \frac{1^3+5(1)^2+3(1)-9}{1-1}=\frac{0}{0}.


This indeterminate form suggests that, we simplify the function first.


We factor to obtain,


\lim_{x \to 1} \frac{(x-1)(x+3)^2}{x-1}


We cancel common factors to get,


\lim_{x \to 1} (x+3)^2


=(1+3)^2=16


The correct answer is D



QUESTION 4

We can see from the table that as x approaches -2 from both sides, the function approaches -4


Hence the limit is -4.


See attachment


The correct answer is option A

3 0
3 years ago
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