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antoniya [11.8K]
2 years ago
15

Someone help me this is due tm

Mathematics
2 answers:
grandymaker [24]2 years ago
8 0

Answer:a I think although I’m really tired

Step-by-step explanation:

Anestetic [448]2 years ago
3 0
AB=VB is the answer hope it helps btw hi
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Find the solution(s) for x in the equation below.
Kobotan [32]

Answer:

D) x=5, x=-5.

Step-by-step explanation:

x^2-25=0

factor

(x+5)(x-5)=0

x+5=0, x-5=0,

x=0-5=-5,

x=0+5=5.

5 0
3 years ago
I need help with Q11 ASAP!!!
storchak [24]

Answer:

Hey there!

When z=48, y=6.

z is 8 times as much as y.

11 part 1: z=8y

11 part 2: I'm going to attach a graph I created at the bottom of this answer :)

Let me know if this helps :)

3 0
3 years ago
Last year the backgammon club had 30 members this year the club has 24 members find the percent of decrease of members
olchik [2.2K]

Answer:

Percent Decrease = 20% Decrease

Step-by-step explanation:

To find percentage decrease, we will use a simple formula shown below:

Percentage Decrease/Increase = \frac{New-Old}{Old}*100

Where

New is the latest number of members (which is 24)

Old is the initial number of members (which is 30)

<em>Note: the multiplication by a factor of "100" brings the answer to a percentage</em>.

Now, lets substitute the numbers and figure out the percentage decrease:

\frac{New-Old}{Old}*100\\\frac{24-30}{30}*100\\\frac{-6}{30}*100\\-0.2*100\\-20

So, it comes to -20, which is a decrease of 20%

7 0
3 years ago
If a cross-country runner runs a 55 kilometer (km) race, how many miles is that? Round your answer to one decimal place.
prohojiy [21]

Answer: 34.2

For everyone 1 kilometer=.621371 so if you times that by 55 you get 34.1754 which rounds to 34.2

5 0
3 years ago
The first three terms of a sequence are given. Round to the nearest thousandth (if
Mumz [18]

Answer:

\frac{3}{16}

Step-by-step explanation:

The common ratio is \frac{1}{2}

24, 12, 6, 3, \frac{3}{2}, \frac{3}{4}, \frac{3}{8}, \frac{3}{16}

so divide each term by 2

<em>plz mark m brainliest. :)</em>

6 0
3 years ago
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