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Olenka [21]
3 years ago
11

The area of the rectangle flower bed shown is

Mathematics
1 answer:
sp2606 [1]3 years ago
6 0

Answer:

I think 5.4

Step-by-step explanation:

Take it and divide by the number of sides

You might be interested in
123
ser-zykov [4K]

First, I'm going to separate factor the expression inside of the square root.

sqrt[ (2/18) * (x^5) ]

sqrt[ (1/9) * (x^5) ]

We can take the square root of 1/9 easily, because 1 and 9 are both perfect squares. The square root of 1/9 is 1/3.

Looking at the x^5, we can separate it into x^2 * x^2 * x^1. The square root of x^2 is x. So, we now also have an x^2 on the outside because there are two x^2s in our expanded form.

ANSWER: (x^2 * sqrt(x)) / 3

(Option 1)

Hope this helps!

4 0
3 years ago
Can someone please help me thank you
Step2247 [10]
Pick one coordinate point on figure 1 (-4,5)

First rotate 90 degree clockwise rule (x,y) -->(y , -x)
So (-4,5) rotate 90 degree clockwise will become (5, 4)

Then reflection over x a-xis, rule (x,y) -->(x,-y)

So (5, 4) reflection over x a-xis will be (5, -4)

Answer: first option
90 clockwise rotation around the origin, then a reflection across x - axis.

Hope it helps
6 0
3 years ago
Find the highest common factor using prime factorization for 27y and 54y³<br>​
amid [387]

Given:

27y and 54y³

​To find:

The highest common factor (HCF) using prime factorization.

Solution:

We have,

27y and 54y³

Using the prime factorization, we get

27y=3\cdot 3\cdot 3\cdot y

54y^3=2\cdot 3\cdot 3\cdot 3\cdot y\cdot y\cdot y

Now, the common prime factors are 3, 3, 3 and y. So,

HCF=3\cdot 3\cdot 3\cdot y

HCF=27y

Therefore, the highest common factor of the given terms is 27y.

5 0
3 years ago
26 is 80% of what number
Tanya [424]
The closest answer I got was 32
5 0
3 years ago
Given: KLIJ is inscribed in circle k(O)
Tems11 [23]

Check the picture below.

let's notice that the angle at K is an inscribed angle with an intercepted arc

\bf \stackrel{\textit{using the inscribed angle theorem}}{K=\cfrac{\widehat{LI}+\widehat{IJ}}{2}}\implies 9x+1=\cfrac{(10x-1)+59}{2} \\\\\\ 9x+1=\cfrac{10x+58}{2}\implies 18x+2=10x+58\implies 8x+2=58 \\\\\\ 8x=56\implies x=\cfrac{56}{8}\implies x=7 \\\\[-0.35em] ~\dotfill\\\\ K=9x+1\implies K=9(7)+1\implies \boxed{K=64}

now, let's notice something again, the angle at L is also an inscribed angle, intercepting and arc of 97 + 59 = 156, so then, by the inscribed angle theorem,

∡L is half that, or 78°.

now, let's take a look at the picture down below, to the inscribed quadrilateral conjecture, since ∡J and ∡I are both supplementary angles, then

∡I = 180 - 64 = 116°.

∡J = 180 - 78 = 102°.

7 0
3 years ago
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