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alexandr1967 [171]
2 years ago
6

Fill in the blanks. If a quadrilateral is both a __ and a __ , then it is a square.

Mathematics
2 answers:
sp2606 [1]2 years ago
7 0

Answer:

rhombus and parralelogram

Step-by-step explanation:

Elodia [21]2 years ago
6 0

Answer: rhombus and parralelogram

Step-by-step explanation:

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Which sequence begins with 1 and follows this rule? multiply by 2 and add 1.
Olin [163]
Change the information into an equation and do sequence. pull out any necessary information.
y = 2x + 1
starts at 1... so your first sequence is 1.
plug in x=1 then it becomes 3. then the next sequence you plug in 3 (, the answer to your previous solve) you get 7. continue to do this. next sequence you plug in x=7 into the equation y = 2x + 1.
thus, y = 2(7) + 1 =15... so forth...
therefore the answer is B.

Hope this helps :)
5 0
3 years ago
n the figure below, A, B, and C represent the areas of the squares with side lengths a, b, and c, respectively. If B = 441 squar
atroni [7]

Answer:

hello 765

Step-by-step explanation:

3 0
3 years ago
What is the GCF of 28a^3 b and 16ab ? <br><br> A)4ab<br> B)4a3b<br> C)28ab<br> D)112a3b2
N76 [4]
28a⁴b: 2 · 2 · 7 · a · a · a · a · b
 16ab: 2 · 2 · 2 · 2 · a · b

GCF(28a⁴b, 16ab) = 4ab

The answer is A.
7 0
3 years ago
What is the diameter of the following circle?<br>​
DochEvi [55]

Answer:

12

Step-by-step explanation:

diameter is 2r

r=6

6*2=12

6 0
3 years ago
Read 2 more answers
The diagram shows the sector of a circle with the centre O and radius 6cm.
cricket20 [7]

Answer:

Area of the shaded region = 1.92 cm²

Step-by-step explanation:

From the picture attached,

Area of the shaded region = Area of the sector OMN - Area of the triangle OMN

Area of sector OMN = \frac{\theta}{360}(\pi r^{2})

Here, θ = Central angle of the sector

r = radius of the sector

Area of sector OMN = \frac{50}{360}(\pi )(6)^2

                                  = 15.708 square cm

Area of ΔOMN = 2(ΔOPN)

Area of ΔOPN = \frac{1}{2}(OP)(PN)

Area of ΔOMN = OP × PN

In ΔOPN,

sin(25°) = \frac{PN}{ON}

PN = ONsin(25°)

     = 6sin(25°)

     = 2.536 cm²

cos(25°) = \frac{OP}{ON}

OP = ONcos(25°)

OP = 6cos(25°)

OP = 5.438 cm

Area of ΔOMN = 2.536 × 5.438

                         = 13.791 cm²

Area of the shaded region = 15.708 - 13.791 = 1.917 cm²

                                             ≈ 1.92 cm²

3 0
3 years ago
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