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3 years ago

I WILL MARK BRAINLEST IF YOU HELP ME

Mathematics

1 answer:

muminat3 years ago

8 0

Answer:

Step-by-step explanation:

top left r= 0.9 ; top right r= 0.4

bottom left r=0; bottom right r= -0.4

if they are not around a line looking like r=0

if they resemble a line well is close to r=1

if line goes up from left to right is r positive

if line goes down from left to right is r negative

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I need the answers please if you can answer all ur a beast

VikaD [51]

Answer:

2. = 24

3. = 6

4. = 30

5. = 15

6. = 8

7. = 40

8. = 14

9. = 60

10. = 9

11. = 40

3 0

3 years ago

Help guys i don’t understand

pishuonlain [190]

Answer:

Step-by-step explanation:

You need to set the problem up like this:

VX=VW+WX. 4x=3x+5

subtract 3 from both sides. x=5

because 'x' equals 5, plug that 4(5)=3(5)+5

back in for all the 'x's

simplify the equation 20=20
The length of the line segment

should be 20

4 0

1 year ago

pls help i have been doing this for 5 hours i will mark brainliest PLEASE I JUST NEED THE RIGHT ANSWER

jenyasd209 [6]

Answer:

$p = 7$

Step-by-step explanation:

So the key is that a line has a measure of 180 degrees. Notice the square between the rays. That is a 90-degree angle. So let's add this up and see what's going on

$(5p + 2) + 90 + (7p + 4) = 180$

$12p + 96 = 180$

$12p = 84$

$p = 7$

6 0

3 years ago

Read 2 more answers

Part A

Vinvika [58]

Answer:

234

Step-by-step explanation:

3 0

3 years ago

Please help me right away with these problems.

viktelen [127]

Q.34

$\sum\limits_{k=1}^{\infty}420\left(1.002\right)^{k-1}$

The infinite geometric series is converges if |r| < 1.

We have r =1.002 > 1, therefore our infinite geometric series is Diverges

Answer: c. Diverges, sum not exist.

Q.35

$\sum\limits_{k=1}^{\infty}-5\left(\dfrac{4}{5}\right)^{k-1}$

The infinite geometric series is converges if |r| < 1.

We have r = 4/5 < 1, therefore our infinite geometric series is converges.

The sum S of an infinite geometric series with |r| < 1 is given by the formula :

$S=\dfrac{a_1}{1-r}$

We have:

$a_1=-5\left(\dfrac{4}{5}\right)^{1-1}=-5\left(\dfrac{4}{5}\right)^0=-5\\\\r=\dfrac{4}{5}$

substitute:

$S=\dfrac{-5}{1-\frac{4}{5}}=-\dfrac{5}{\frac{1}{5}}=-5\cdot\dfrac{5}{1}=-25$

Answer: c. Converges, -25.

4 0

3 years ago