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grandymaker [24]
3 years ago
11

Make a ten or a hundred to add mentally. 198 + 132 1,274 + 3,599

Mathematics
1 answer:
Svet_ta [14]3 years ago
8 0
198+ 132= 330
1274+ 3599= 4873
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20 POINTS+BRAINLIEST: 8TH GRADE MATH PART 3: Word problem with fractions - please list out the steps and explain how you came to
Greeley [361]

Answer: H. 12

Step-by-step explanation:

Add all cost of each price of the each fabric once to get $925 (150+275+500). Then, divide 5,550/925 to get 6. Then multiply 1/3 by 6 to get 2. Then multiply 2/3 by 6 to get 4. Then add 2+4+6 to get 12.

5 0
1 year ago
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Three and seventt-two hundredths subtracted from five eighty-one hundredths in standard form
labwork [276]
5.81 -  3.72 = 2.09 

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4 0
4 years ago
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Find the area and perimeter of ABC at right. Give approximate (decimal) answers, not exact answers
AURORKA [14]

Answer:

Area of Δ ABC = 21.86 units square

Perimeter of Δ ABC = 24.59 units

Step-by-step explanation:

Given:

In Δ ABC

∠A=45°

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To find area and perimeter of triangle we need to find the sides of the triangle.

Naming the end point of altitude as 'O'

Given BO\perp AC

For Δ ABO

Since its a right triangle with one angle 45°, it means it is a special 45-45-90 triangle.

The sides of 45-45-90 triangle is given as:

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∴ AB (hypotenuse) =x\sqrt2=4\sqrt2=5.66  

For Δ CBO

Since its a right triangle with one angle 30°, it means it is a special 30-60-90 triangle.

The sides of 30-60-90 triangle is given as:

We are given BO (side opposite 30° angle) =x=4

CO (side opposite 60° angle) =x\sqrt3=4\sqrt3=6.93

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AC=AO+CO=4+6.93=10.93

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⇒ AB+BC+AC

⇒ 5.66+8+10.93

⇒ 24.59 units

Area of Δ ABC = \frac{1}{2}\times base\times height

⇒  \frac{1}{2}\times 10.93\times 4

⇒ 21.86 units square

6 0
3 years ago
Write the sentence as an equation.<br> z plus 372 is equal to 160
Makovka662 [10]

Answer:

z+372=160

Step-by-step explanation:

7 0
4 years ago
Bonjour pouvez vous m'aider pour les exercices 81 à 85 je suis en 6ème. Merci de m'aider.
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Je vous aiderais... mais y a-t-il une question pour elle ?
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3 years ago
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