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SashulF [63]
3 years ago
11

Please help me with this. I need a maths genius See pic below. Thanks

Mathematics
1 answer:
adelina 88 [10]3 years ago
5 0

Answer:

a =15

b=70

c=400

d=1500

e=5300

Step-by-step explanation:

it will help u

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Alma got to the playground at 2:45.she spent 20 minutes on the swings and 10 minutes on the jungle gym. She played on the slide
Archy [21]

Answer:

Alma got home at 3:27

Step-by-step explanation:

2:45+30 min=3:15

3:15+12 min = 3:27

3 0
3 years ago
Solve the inequality.<br> 4│x + 5│ - 2 ≤ 10
castortr0y [4]

Step-by-step explanation: To solve this absolute value inequality,

our goal is to get the absolute value by itself on one side of the inequality.

So start by adding 2 to both sides and we have 4|x + 5| ≤ 12.

Now divide both sides by 3 and we have |x + 5| ≤ 3.

Now the the absolute value is isolated, we can split this up.

The first inequality will look exactly like the one

we have right now except for the absolute value.

For the second one, we flip the sign and change the 3 to a negative.

So we have x + 5 ≤ 3 or x + 5 ≥ -3.

Solving each inequality from here, we have x ≤ -2 or x ≥ -8.

6 0
3 years ago
Read 2 more answers
Which function is undefined for x = 0? y=3√x-2 y=√x-2 y=3√x+2 y=√x=2
Mkey [24]

For this case, we have to:

By definition, we know:

The domain of f (x) = \sqrt [3] {x} is given by all real numbers.

Adding or removing numbers to the variable within the root implies a translation of the function vertically or horizontally. In the same way, its domain will be given by the real numbers, independently of the sign of the term inside the root. Thus, it will always be defined.

So, we have:

y = \sqrt [3] {x-2} withx = 0: y = \sqrt [3] {- 2} is defined.

y = \sqrt [3] {x+2}with x = 0:\ y = \sqrt [3] {2} is also defined.

f (x) = \sqrt {x}has a domain from 0 to ∞.

Adding or removing numbers to the variable within the root implies a translation of the function vertically or horizontally. For it to be defined, the term within the root must be positive.

Thus, we observe that:

y = \sqrt {x-2} is not defined, the term inside the root is negative whenx = 0.

While y = \sqrt {x+2} if it is defined for x = 0.

Answer:

y = \sqrt {x-2}

Option b

6 0
3 years ago
Scott rented a truck for a day there was a base fee of $14.95 and there was an additional charge of $0.83 for each mile driven S
Mademuasel [1]

Answer:

288

Step-by-step explanation:

1. Subtract 14.95 from 253.99

2. Then you are going to divide 239.04 by 0.83

3. Your answer will be 288

4 0
3 years ago
PLEASE HELP WILL GIVE OUT BRAINLEYST<br> MATHH PEOPLE HELP
Umnica [9.8K]

Answer its 82 i added the cm

3 0
3 years ago
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