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Lena [83]
3 years ago
7

1.The diagonals of a rhombus are perpendicular.. . always sometimes. never. . 2.The diagonals of a rectangle are equal.. . alway

s sometimes. never. . 3.The diagonals of a trapezoid are perpendicular.. . always. sometimes -and this. never.
Mathematics
2 answers:
melisa1 [442]3 years ago
5 0

the first two are always the last one is sometimes

Hoochie [10]3 years ago
3 0
1. Answer – Always
The diagonals of a rhombus are always perpendicular. In fact, if the diagonals of a parallelogram are perpendicular bisectors of each other, then it must be a rhombus. In addition to this, a rhombus always has all four congruent sides.
 2. Answer – Always
The diagonals of a rectangle are always congruent (i.e. they are equal in length). Furthermore, in every rectangle, the diagonals bisect each other. Because the diagonals are equal in length, each diagonal divides the rectangle into two congruent right triangles.
 3. Answer – Never
The diagonals of a trapezoid are never perpendicular; the diagonals are only perpendicular in rhombuses and squares. In a trapezoid however, the diagonals are congruent. A trapezoid is a quadrilateral that has only one pair of parallel sides. <span>
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Please help, will mark brainliest .
yan [13]

18/3=6

24/3=8

21/3=7

The missing side is 7

4 0
3 years ago
Show with work please.
kolbaska11 [484]

Answer:

$\csc \left(\theta-\frac{\pi }{2}\right)=0.73$

Step-by-step explanation:

The identity you will use is:

$\csc \left(x\right)=\frac{1}{\sin \left(x\right)}$

So,

$\csc \left(\theta-\frac{\pi }{2}\right)$

$\csc \left(\theta-\frac{\pi }{2}\right)=\frac{1}{\sin \left(-\frac{\pi }{2}+\theta\right)}$

Now, using the difference of sin

Note: state that \text{sin}(\alpha\pm \beta)=\text{sin}(\alpha) \text{cos}(\beta) \pm \text{cos}(\alpha) \text{sin}(\beta)

$\csc \left(\theta-\frac{\pi }{2}\right)=\frac{1}{-\cos \left(\theta\right)\sin \left(\frac{\pi }{2}\right)+\cos \left(\frac{\pi }{2}\right)\sin \left(\theta\right)}$

Solving the difference of sin:

$-\cos \left(\theta\right)\sin \left(\frac{\pi }{2}\right)+\cos \left(\frac{\pi }{2}\right)\sin \left(\theta\right)$

-\cos \left(\theta\right) \cdot 1+0\cdot \sin \left(\theta\right)

-\text{cos} \left(\theta\right)

Then,

$\csc \left(\theta-\frac{\pi }{2}\right)=-\frac{1}{\cos \left(\theta\right)}$

Once

\text{sec}(-\theta)=\text{sec}(\theta)

And, \text{sec}(\theta)=-0.73

$-\frac{1}{\cos \left(\theta\right)}=-\text{sec}(\theta)$

$-\frac{1}{\cos \left(\theta\right)}=-(-0.73)$

$-\frac{1}{\cos \left(\theta\right)}=0.73$

Therefore,

$\csc \left(\theta-\frac{\pi }{2}\right)=0.73$

3 0
3 years ago
Nancy is going to glaze 6-ounce servings of pretzels and put them into gift boxes. If pretzels
Shtirlitz [24]
When in doubt guess C. Always works for me
8 0
3 years ago
Is 11.250 greater or less or equal to 11.25
sasho [114]

Answer:

Equal, 250 and 25 are both 1/4

Step-by-step explanation:

4 0
4 years ago
Read 2 more answers
HELPPPPP PLEASSEEEEEEE
NikAS [45]
The answer would be A.132 because when you add the corner length to 90 it equals 132
7 0
3 years ago
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