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Aleks [24]
3 years ago
14

The length of a square is 10 cm. Calculate the perimeter of the square A. 40cm² B. 400cm² C. 4ocm D. 414cm² I will mark you as b

rainliest
Mathematics
2 answers:
Studentka2010 [4]3 years ago
4 0

Answer:

Hey !! Side of square = 10 cm. Therefore, Perimeter of square = 4 × ( Side ) = 4 × 10 = 40 cm

I holpe it helps pleas mark me

Step-by-step explanation:

noname [10]3 years ago
3 0

Answer:

C. 40 cm

Step-by-step explanation:

length of side of square = s

perimeter = s + s + s + s = 4s

We use the formula P = 4s

P = 4 * 10 cm

P = 40 cm

Answer: The perimeter is 40 cm.

Notice that the perimeter is a sum of lengths, so its units are linear units such as cm, inches, feet, a unit of length.

An area has square units such as cm^2, in.^2, ft^2, etc.

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The height of the tree is 3m

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How much lighter is 345 g than 1,1 kg​
NemiM [27]

Answer:

755 gram lighter

Step-by-step explanation:

Convert 1.1kg into gram by multiplying it with 1000

1.1 x 1000= 1100 and we take this amount and subtract it with 345g

1100-345= 755g

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3 years ago
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valentinak56 [21]
4x-4>20

Step 1: Add four to both sides
4x-4+4>20+4
4x>24
Step 2: Divide both sides by 4
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788.53 expressed to the neareast whole number​
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789

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Step-by-step explanation:

8 0
3 years ago
Jane wishes to bake an apple pie for dessert. The baking instructions say that she should bake the pie in an oven at a constant
Viktor [21]

Answer:

Therefore k= \frac{ln2 }{18}, A=184

Step-by-step explanation:

Given function is

T(t)=230 -e^{-kt}

where T(t) is the temperature in °C and t is time in minute and A and k are constants.

She noticed that after 18 minutes the temperature of the pie is 138°C

Putting T(t) =138°C and t= 18 minutes

138=230 -Ae^{-k\times 18}

\Rightarrow  -Ae^{-18k}=138-230

\Rightarrow  Ae^{-18k}=92 .....(1)

Again after 36 minutes it is 184°C

Putting T(t) =184°C and t= 36 minutes

184=230-Ae^{-k\times 36}

\Rightarrow Ae^{-36k}=230-184

\Rightarrow Ae^{-36k}=46.......(2)

Dividing (2) by (1)

\frac{Ae^{-36k}}{Ae^{-18k}}=\frac{46}{92}

\Rightarrow e^{-18k}=\frac{46}{92}

Taking ln both sides

ln e^{-18k}=ln\frac{46}{92}

\Rightarrow -18k =ln (\frac12)

\Rightarrow -18k= ln1-ln2

\Rightarrow k= \frac{ln2 }{18}

Putting the value k in equation (1)

Ae^{-18\frac{ln2}{18}}=92

\Rightarrow A e^{ln2^{-1}}=92

\Rightarrow A.2^{-1}=92

\Rightarrow \frac{A}{2}=92

\Rightarrow A= 92 \times 2

⇒A= 184.

Therefore k= \frac{ln2 }{18}, A=184

7 0
3 years ago
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