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PIT_PIT [208]
3 years ago
15

Use the parabola tool to graph the quadratic function f(x) =-5x2 -2

Mathematics
1 answer:
enot [183]3 years ago
3 0
Your first point would be on (0, 2), the vertex. Then, put a point on (-2, 22). Hope that helps!
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HELLPPPP I WILL GIVE BRAINIEST!!!!!!!!!!!!<br><br> 2x+3=20-15
Nesterboy [21]

Here you go!

Answer: X = 1



8 0
3 years ago
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Helppppppppppppp i need this homework done by today
lbvjy [14]

Answer: 13. -68

14. 36

15. -36

16. -3

17. 10

18. -4

19. 1310 ft

20. 330 ft

Step-by-step explanation:

13. 34*(-2)=-68

14. -9*(-4)=

-(9*(-4))=

-(-36)=36

15. 12*(-3)=-36

16. -12/4=-3

17. -20/(-2)=

20/2=10

18. 200/(-50)=-4

19. -1640*(-1)-(-330*(-1))=

1640-(-330*(-1))=

1640-330=1310 ft

20. -990*(-1)-(-660*(-1))=

990-(-660*(-1))=

990-660=330 ft

7 0
1 year ago
An elevator descends into a mine shaft at the rate of 6 m/min. If the descend starts from 20 meter above the ground level, how l
levacccp [35]
60 minutes. Take how far it is above ground (20 m) and add it to the absolute value of how far it is below ground (340 m) to get 360 meters, which you divide by the rate of descent (6 m/min) to get 60 minutes.
5 0
3 years ago
Matthew invested $3,000 into two accounts. One account paid 3% interest and the other paid 8% interest. He earned 4% interest on
boyakko [2]

<u>Answer:</u>

<em>Mathew invested</em><em> $600 and $2400</em><em> in each account.</em>

<u>Solution:</u>

From question, the total amount invested by Mathew is $3000. Let p = $3000.

Mathew has invested the total amount $3000 in two accounts. Let us consider the amount invested in first account as ‘P’

So, the amount invested in second account = 3000 – P

Step 1:

Given that Mathew has paid 3% interest in first account .Let us calculate the simple interest (I_1) earned in first account for one year,

\text {simple interest}=\frac{\text {pnr}}{100}

Where  

p = amount invested in first account

n = number of years  

r = rate of interest

hence, by using above equation we get (I_1) as,  

I_{1}=\frac{P \times 1 \times 3}{100} ----- eqn 1

Step 2:

Mathew has paid 8% interest in second account. Let us calculate the simple interest (I_2) earned in second account,

I_{2} = \frac{(3000-P) \times 1 \times 8}{100} \text { ------ eqn } 2

Step 3:

Mathew has earned 4% interest on total investment of $3000. Let us calculate the total simple interest (I)

I = \frac{3000 \times 1 \times 4}{100} ----- eqn 3

Step 4:

Total simple interest = simple interest on first account + simple interest on second account.

Hence we get,

I = I_1+ I_2 ---- eqn 4

By substituting eqn 1 , 2, 3 in eqn 4

\frac{3000 \times 1 \times 4}{100} = \frac{P \times 1 \times 3}{100} + \frac{(3000-P) \times 1 \times 8}{100}

\frac{12000}{100} = \frac{3 P}{100} + \frac{(24000-8 P)}{100}

12000=3P + 24000 - 8P

5P = 12000

P = 2400

Thus, the value of the variable ‘P’ is 2400  

Hence, the amount invested in first account = p = 2400

The amount invested in second account = 3000 – p = 3000 – 2400 = 600  

Hence, Mathew invested $600 and $2400 in each account.

3 0
3 years ago
Read 2 more answers
I need help with this question
Lunna [17]
{3,4,5,6} should be the answer
8 0
3 years ago
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