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3 years ago

Please help me with these questionsAny help will be appreciated.

Mathematics

2 answers:

satela [25.4K]3 years ago

7 0

Answer:

Step-by-step explanation:

28) Perimeter = 10 + 2 + (10 -2) + 3 + (8-2) + 1 + 8 + (1+3+2)

= 10 + 2 + 8 + 3 + 6 + 1 + 8 + 6

= 44 cm

29) Area = Area of upper block + area of middle block + area of the down block

= (10 * 2) + (2*3) + (1*8)

= 20 + 6 + 8

= 34 cm²

saul85 [17]3 years ago

4 0

Answer:

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Step-by-step explanation:

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Mars2501 [29]

Multiply both sides by 30
30(15/18) = y
we then multiply 30*15 = 450
450/18 = y
we then get
y = 25

5 0

3 years ago

an elderly patient goes to a general doctor every 4 weeks and goes to a chiroprator every 6 weeks . if he went to both the gener

Fed [463]

Answer:

Step-by-step explanation:

He will actually need to wait for 12 weeks. And that is because if you look at the multiples of 6 and 4 and you'll notice that 12 is the lowest common multiple between them. So you start counting from 6 and 4 at the same time until you notice the LCM of these two numbers then on which ever you number you settle on is how long he will need to wait.

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3 years ago

20 POINTS AND BRAINLIEST

emmainna [20.7K]

To solve this equation we must first plug in 5 for x.

$\frac{1}{3}$ <span> (5 - 4x)
</span>↓
$\frac{1}{3}$ (5 - 4 · 5)

The next step would be to multiply 4 by 5 to get rid of the parenthesis within our main parenthesis.

$\frac{1}{3}$ (5 - 4 · 5)

$\frac{1}{3}$ (5 - 20)

The next step would be to subtract both of the numbers within the parenthesis.

$\frac{1}{3}$ (-15)

Now we must multiply.

- $\frac{1}{3}$ · 15 = - $\frac{1×15}{3}$

- $\frac{15}{3}$

Now we simplify our answer to get our final answer.

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3 years ago

The question in picture attached

krek1111 [17]

Need free pointtdsssss

4 0

3 years ago

I need the answer to this please.

ohaa [14]

Answer:

Step-by-step explanation:

Its (1,3) and (-3, -2)

6 0

3 years ago

Please help me with these questionsAny help will be appreciated.​

Please help me with these questionsAny help will be appreciated.