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3 years ago

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

3 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