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3 years ago

Parallelogram ABCD has the angle measures shown. Can you conclude that it is a rhombus, a rectangle, or a square? Explain.

Mathematics

2 answers:

Nikitich [7]3 years ago

7 0

<span>We can analyze the four optons. 1) Option A. A parallelogram with all four angles of the same measure can be either a square or a rectangle, then this option is not valid. 2) Optrion B. gives not information. 3) A rhombus (a diamond) is a parallelogram with four congruent side (square is a specific case of rhmbus but not all rhombus are squares), and it is enouh to say that one diagonal bisects two interior angles, to conclude that it is a rhombus. 4) If a diagonal creates congruent angles, but you do not know what happens with the opposed angle, you cannot conclude that the parallelogram is a rectangle; it could be a trapezoid with one side perpendicular to the parallel sides. By t his analysis, the answer is option C.</span>

Citrus2011 [14]3 years ago

5 0

Answer:

Step-by-step explanation:

Refer the attached figure :

In Δ ABD ,

∠B=∠D = 66°

By property : opposite sides of equal angles are equal

Thus AB=AC

In Δ CBD ,

∠B=∠D = 66°

By property : opposite sides of equal angles are equal

Thus CB=CD

Thus all four sides of quadrilateral ABCD are equal

And diagonal BD bisects the angles

So, it is a rhombus

Thus Option c is correct .

c. Parallelogram ABCD is a rhombus, because the diagonal bisects two angles.

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blondinia [14]

Answer:

Step-by-step explanation:

Put x = 2 into both equations:

For x- 10 = 2-10 = 8 (Not > 8)

For -0.5x-4 = -0.5*2-4 = -5 (Not < -8)

Hence, the answer is no.

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Step-by-step explanation:

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2 years ago

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3 years ago

How can I solve this?

Eva8 [605]

Answer:

they are congruent so 47 is also the measurement for its corresponding angle(aka the corner that doesn't have a variable)

and in a triangle, interior angles add to 180 so 47+x+y=180

90 + 43 = 133

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so your answer is a

hope this helps!!

8 0

3 years ago

Question 2 of 5

AlekseyPX

Given:

The different recursive formulae.

To find:

The explicit formulae for the given recursive formulae.

Solution:

The recursive formula of an arithmetic sequence is $f(n)=f(n-1)+d, f(1)=a,n\geq 2$ and the explicit formula is $f(n)=a+(n-1)d$ , where a is the first term and d is the common difference.

The recursive formula of a geometric sequence is $f(n)=rf(n-1), f(1)=a,n\geq 2$ and the explicit formula is $f(n)=ar^{n-1}$ , where a is the first term and r is the common ratio.

The first recursive formula is:

$f(1)=5$

$f(n)=f(n-1)+5$ for $n\geq 2$ .

It is the recursive formula of an arithmetic sequence with first term 5 and common difference 5. So, the explicit formula for this recursive formula is:

$f(n)=5+(n-1)(5)$

$f(n)=5+5(n-1)$

Therefore, the correct option is A, i.e., $f(n)=5+5(n-1)$ .

The second recursive formula is:

$f(1)=5$

$f(n)=3f(n-1)$ for $n\geq 2$ .

It is the recursive formula of a geometric sequence with first term 5 and common ratio 3. So, the explicit formula for this recursive formula is:

$f(n)=5(3)^{n-1}$

Therefore, the correct option is F, i.e., $f(n)=5(3)^{n-1}$ .

The third recursive formula is:

$f(1)=5$

$f(n)=f(n-1)+3$ for $n\geq 2$ .

It is the recursive formula of an arithmetic sequence with first term 5 and common difference 3. So, the explicit formula for this recursive formula is:

$f(n)=5+(n-1)(3)$

$f(n)=5+3(n-1)$

Therefore, the correct option is D, i.e., $f(n)=5+3(n-1)$ .

6 0

3 years ago

Read 2 more answers