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AlladinOne [14]
3 years ago
10

Name the quadrilateral whose diagonals are perpendicular bisectors of each other

Mathematics
1 answer:
dexar [7]3 years ago
5 0

Rhombus or square.

Hope it helps!

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Jason rowed 1/3 of a mile down the river.Jane rowed 2/6 of a mile down the river. Jude rowed 2/9 of a mile down the river.What i
seraphim [82]

Answer:

8 / 9 miles

Step-by-step explanation:

Given that:

Jason's distance = 1/3 mile

Jane's distance = 2/6 miles

Jude's distance = 2/9 miles

The total miles rowed down the river by all 3 :

(1/3 + 2/6 + 2/9) miles

The L. C. M of the denominator (3, 6, 9) = 18

(6 + 6 + 4) / 18

16 / 18 miles

8 / 9 miles

4 0
3 years ago
Identify the roots of the equation and the multiplicities of the roots.<br> (x - 6)(x + 4)2 = 0
Harman [31]

Answer:

Roots are x=6 and - 4

Step-by-step explanation:

2(x-6)(x+4)=0

(x-6)=0 or (x+4)=0

x=6 or x=-4, their multiplicities is 1

8 0
3 years ago
Function that has <br> Domain: x≠-1<br> Range: y≠2
Degger [83]
I don't understand your question.
4 0
3 years ago
Heather can complete two problems in ten minutes, and when she started she had three problems done. joel can complete three prob
zzz [600]
 heather can do 2 problems in 10 minutes....60 minutes in an hr
2/10 = x / 60...2 problems to 10 min = x problems to 60 min
10x = 120
x = 120/10
x = 12....so she can do 12 problems in 1 hr

Joel can complete 3 problems in 15 minutes...
3/15 = x / 60
15x = 180
x = 180/15
x = 12....so he can do 12 problems in 1 hr

so 24 problems can be done in 1 hr

Heather started with 3 problems done and Joel started with 2 problems done....added = 5 problems already done

and it is asking between 30 and 50 hrs.....so there is no equal sign in ur problem

ur answer is : 30 < 24x + 5 < 50 <==




5 0
3 years ago
What is the angle measure of an arc bounding sector with an area of 5pie square miles?
Snowcat [4.5K]

if the diameter is 20, the its radius must be half that or 10.

\textit{area of a sector of a circle}\\\\ A=\cfrac{\theta \pi r^2}{360}~~ \begin{cases} r=radius\\ \theta =\stackrel{degrees}{angle}\\[-0.5em] \hrulefill\\ A=5\pi \\ r=10 \end{cases}\implies \begin{array}{llll} 5\pi =\cfrac{\theta \pi (10)^2}{360}\implies 5\pi =\cfrac{5\pi \theta }{18} \\\\\\ \cfrac{5\pi }{5\pi }=\cfrac{\theta }{18}\implies 1=\cfrac{\theta }{18}\implies 18=\theta \end{array}

8 0
3 years ago
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