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lakkis [162]
3 years ago
5

HELPPPP !!! ILL GIVE U BRAINLIEST

Mathematics
2 answers:
Pepsi [2]3 years ago
8 0
I believe the answer is D because those equation indicate an area on the negative side
ahrayia [7]3 years ago
6 0
I agree I think the answer is d
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Helppp!!! i need help
klio [65]

Answer:

rectangle

kite

Step-by-step explanation:

4 0
3 years ago
The diagram shows four graphs match the graphs to the equations
cluponka [151]

Answer:

y=x^2 is graph D

y=1/x is graph A

otherwise, you're right

7 0
2 years ago
Find the sum of a finite geometric sequence from n = 1 to n = 6, using the expression −2(5)n − 1.
vazorg [7]
\bf \qquad \qquad \textit{sum of a finite geometric sequence}\\\\
S_n=\sum\limits_{i=1}^{n}\ a_1\cdot r^{i-1}\implies S_n=a_1\left( \cfrac{1-r^n}{1-r} \right)\quad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
r=\textit{common ratio}\\
----------\\
a_1=-2\\
r=5\\
n=6
\end{cases}
\\\\\\
S_6=-2\left( \cfrac{1-5^6}{1-5} \right)
8 0
3 years ago
Read 2 more answers
Help me to do this please
Tanzania [10]

If x represents the mass of 1 ball in kg, then the balance shows ...

... 6x = x + 9 kg

... 5x = 9 kg

Then 6x is ...

... (6/5)·5x = (6/5)·9 kg = 10.8 kg

0.8 kg = 0.8·1000 g = 800 g

so 10.8 kg is ...

... D. 10 kg 800 g

7 0
3 years ago
Hello I need help Finding the probability
Stolb23 [73]
Each petal of the region R is the intersection of two circles, both of diameter 10. Each petal in turn is twice the area of a circular segment bounded by a chord of length 5\sqrt2, which implies the segment is subtended by an angle of \dfrac\pi2. This means the area of the segment is

\text{area}_{\text{segment}}=\text{area}_{\text{sector}}-\text{area}_{\text{triangle}}
\text{area}_{\text{segment}}=\dfrac{25\pi}4-\dfrac{25}2

This means the area of one petal is \dfrac{25\pi}2-25, and the area of R is four times this, or 50\pi-100.

Meanwhile, the area of G is simply the area of the square minus the area of R, or 10^2-(50\pi-100)=200-50\pi.

So

\mathbb P(X=R)=\dfrac{50\pi-100}{100}=\dfrac\pi2-1
\mathbb P(X=G)=\dfrac{200-50\pi}{100}=2-\dfrac\pi2
\mathbb P((X=R)\land(X=G))=0 (provided these regions are indeed disjoint; it's hard to tell from the picture)
\mathbb P((X=R)\lor(X=G))=\mathbb P(X=R)+\mathbb P(X=G)=1

4 0
3 years ago
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