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Anna71 [15]
3 years ago
5

A. It has point symmetry.

Mathematics
2 answers:
xz_007 [3.2K]3 years ago
8 0

Answer:

C. It has reflectional symmetry with five lines of symmetry.

Step-by-step explanation:

Goryan [66]3 years ago
4 0

Answer: C. It has reflectional symmetry with five lines of symmetry.

The five lines of symmetry are shown in the diagram below. Each line splits a flower petal in half. Each line also splits the entire flower in half. In other words, reflecting one half of a flower over any given line of symmetry, generates the other half of the flower.

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Aurelio can choose between 3 pairs of pants and 4 shirts. How many different outcomes are in the sample space?
kherson [118]

Answer:

12

Step-by-step explanation:

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4 years ago
5. The value of y varies directly with x, and
Kruka [31]

Answer: y= 3x

Step-by-step explanation:

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8 0
3 years ago
For each question below, use the function f (x) = x3
givi [52]

Answer:

A. (-∞ , 0)

B. x = 0 triple root

C. The function tends to infinity

Step-by-step explanation:

For the function f(x) = x^3 we have:

Negative intervals:

(-∞ , 0)

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A graphic is attached below


5 0
3 years ago
Read 2 more answers
Can you prove it??<br>it's hard,I tried but couldn't solve this.
BARSIC [14]

I'll abbreviate s=\sin\theta and c=\cos\theta, so the identity to prove is

\dfrac{s+c+1}{s+c-1}-\dfrac{1+s-c}{1-s+c}=2\left(1+\dfrac1s\right)

On the left side, we can simplify a bit:

\dfrac{s+c+1}{s+c-1}=\dfrac{s+c-1+2}{s+c-1}=1+\dfrac2{s+c-1}

\dfrac{1+s-c}{1-s+c}=-\dfrac{-2+1-s+c}{1-s+c}=-1+\dfrac2{1-s+c}

Then

\dfrac{s+c+1}{s+c-1}-\dfrac{1+s-c}{1-s+c}=2\left(1+\dfrac1{s+c-1}-\dfrac1{1-s+c}\right)

So the establish the original equality, we need to show that

\dfrac1{s+c-1}-\dfrac1{1-s+c}=\dfrac1s

Combine the fractions:

\dfrac{(1-s+c)-(s+c-1)}{(s+c-1)(1-s+c)}=\dfrac{2-2s}{c^2-s^2+2s-1}

We can rewrite the denominator as

c^2-s^2+2s-1=c^2+s^2-2s^2+2s-1

then using the fact that c^2+s^2=\cos^2\theta+\sin^2\theta=1, we get

1-2s^2+2s-1=2s-2s^2

so that we have

\dfrac1{s+c-1}-\dfrac1{1-s+c}=\dfrac{2-2s}{2s-2s^2}=\dfrac1s

as desired.

6 0
4 years ago
I need help with this I don’t understand this
Y_Kistochka [10]
D.1296 is the answer
4 0
3 years ago
Read 2 more answers
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