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puteri [66]
3 years ago
8

1. -30 and 25 2. -34 and -37 3. -15 and 15

Mathematics
2 answers:
ankoles [38]3 years ago
4 0

Answer:

1. -5

2.-71

3. 0

if you think by this im not sure

inessss [21]3 years ago
3 0
What’s the question ???
You might be interested in
Sharon needs 64 credits to graduate from her community college. So far she has earned 56 credits. What percent of the required c
Anit [1.1K]
Sharon has a percentage of 56/64*100= 87.5%
3 0
3 years ago
Read 2 more answers
Chandra created a budget matrix based on her regular and expected expenses for the year. Expense Jan. Feb. Mar. Apr. May June Ju
Ivan

Answer:

Chandra's Average Monthly Expenses are;

1) For Both Jan and Feb are $269

2) For the remaining 10 months each are $244

Step-by-step explanation:

From Chandra's matrix all monthly expenses are all same that is,

Cell phone $71, Rent $1,025, Gym $75, Internet $25, Auto insurance $425, Gas $ 120, Food $145. which are all the expenses carried out every month for 12 months.

That means Chandra carries out a total of 7 expenses for the month of January and February while she carried a total of 6 expenses for the remaining 10 month which was noted from the matrix that Auto insurance was carried out only in the month of January and February.

Therefore, you start by adding up each month total expenses, which are ;

January = $71 + $1,025 + $75 + $25 + $425 + $120 + $145 = $1886

February = $71 + $1,025 + $75 + $25 + $425 + $120 + $145 = $1886

March = $71 + $1,025 + $75 + $25  + $120 + $145 = $1461

April = $71 + $1,025 + $75 + $25  + $120 + $145 = $1461

May = $71 + $1,025 + $75 + $25  + $120 + $145 = $1461

June = $71 + $1,025 + $75 + $25  + $120 + $145 = $1461

July= $71 + $1,025 + $75 + $25  + $120 + $145 = $1461

August = $71 + $1,025 + $75 + $25  + $120 + $145 = $1461

September = $71 + $1,025 + $75 + $25  + $120 + $145 = $1461

October = $71 + $1,025 + $75 + $25  + $120 + $145 = $1461

November = $71 + $1,025 + $75 + $25  + $120 + $145 = $1461

December = $71 + $1,025 + $75 + $25  + $120 + $145 = $1461

Therefore Chandra's Average Monthly Expenses are:

1) January & February  = \frac{71 + 1,025 + 75 + 25 + 425 + 120 + 145}{7} = \frac{1886}{7} = 269.4286 to the nearest cent

≅ $269

2) For the remaining 10 month are = \frac{71 + 1,025 + 75 + 25  + 120 + 145}{6} = \frac{1461}{6}= 243.5 to the nearest cent

≅ $244

8 0
3 years ago
Read 2 more answers
What is the 9th term in the sequence 2, -6, 18, -54, ...?
gregori [183]

Your answer would be 13,122

You would multiply by -3 each time for this particular sequence.

Hope I helped!

6 0
3 years ago
Courtney had $55 to spend at the fair. She paid $22.50 for rides and $14.25 for lunch and snacks. Courtney says she has $18.25 l
IrinaVladis [17]
She has $55. She spent $22.50 on rides so now she has $32.50 . Then she spends $14.25 on lunch and snack so now she has $18.25.


A. would be wrong because Courtney IS correct, she has $18.25 left.

C. is also wrong because, again, Courtney IS correct.

So the only option left is B and it explains why Courtney is correct.

Option B is the correct answer.

Hope this helps.

4 0
3 years ago
For the problem, use the discriminant to determine the number of real solutions for the equation. Then, find the solutions and c
Ksivusya [100]

Answer:

Discriminant = 55.2 > 0 -> 2 real solutions

Solutions: t1 = -1.1663 s and t2 = 0.35 s

The solution t1 doesn't make sense for this problem, as we can't have a negative value for the time.

So the solution is t2 = 0.35 s

Step-by-step explanation:

To find the time when the ball will reach the height of 2 meters, we just need to use the value of h = 2 in the equation given. So, we have that:

−4.9t^2 − 4t + 4 = 2

−4.9t^2 − 4t + 2 = 0

For this equation, we have the constants a = -4.9, b = -4 and c = 2. So the discriminant Delta is:

Delta = b^2 - 4ac = 16 + 39.2 = 55.2

sqrt(Delta) = 7.4297

As Delta > 0, we have 2 real solutions

t1 = (-b + sqrt(Delta)) / 2a = (4 + 7.4297) / (-9.8)  = -1.1663 s

t2 = (-b - sqrt(Delta)) / 2a = (4 - 7.4297) / (-9.8)  = 0.35 s

Number of real solutions: 2

Solutions: t1 = -1.1663 s and t2 = 0.35 s

The solution t1 doesn't make sense for this problem, as we can't have a negative value for the time.

So the solution is t2 = 0.35 s

8 0
3 years ago
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