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Kaylis [27]
3 years ago
6

1000X 35000+6000,000

Mathematics
1 answer:
sammy [17]3 years ago
5 0

Answer:

Multiply  

35000  by  1000 .

35000000 X  +  6000 , 0

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a number has a 4 in the thousands place the number also has a digit whose value is 10 times the value of the 4 on the thousands
maks197457 [2]
Well I can still answer that question without the following and it will be 44,000
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3 years ago
Jada purchases a sweater for $36. The store charges her 8% sales tax on the purchase. How much sales tax will the store charge o
dusya [7]

Answer:

Step-by-step explanation:

36/100=0.36 - 1%

8*0.36=2.88$

6 0
2 years ago
provide a counterexample (specific values of a,b, etc. which make the statement false) for each of the following statements. Ass
ivolga24 [154]

Answer:

1. a=3, b = 2, c = 1.

2. a=2, b=3, c=4, d=6.

3. a = 9, b = 3.

Step-by-step explanation:

1.  3 | 21   is a counterexample  because 3 does not divide into  2 or 1.

2. 23 | 46 is a counterexample because 2 does not divide into 3.

3. 9 | 3^2 is a counterexample because 9 does not divide into 3.

8 0
2 years ago
A triangle with sides 11m , 13m and 18m is a right triangle.<br> ​<br> A<br> True<br> B<br> False
Julli [10]

\bold{\huge{\pink{\underline{ Solution}}}}

\bold{\underline{ Given :-}}

  • <u>A </u><u>triangle </u><u>with </u><u>sides </u><u>11m</u><u>, </u><u> </u><u>13m </u><u>and </u><u>18m</u>

\bold{\underline{ To \: Find :-}}

  • <u>We</u><u> </u><u>have </u><u>to </u><u>check </u><u>it </u><u>whether </u><u>it </u><u>is </u><u>right </u><u>angled </u><u>triangle </u><u>or </u><u>not</u><u>? </u>

\bold{\underline{ Let's \: Begin :-}}

\sf{\red{ In \:right \:angled\ : triangle, }}

According to the Pythagoras theorem, The sum of the squares of perpendicular height and the square of the base of the triangle is equal to the square of hypotenuse that is sum of the squares of two small sides equal to the square of longest side of the triangle.

<u>We </u><u>imply</u><u> </u><u>it </u><u>in </u><u>the </u><u>given </u><u>triangle </u><u>,</u>

\sf{\red{ ( Perpendicular)² + (Base)² = (Hypotenuse)²}}

\sf{(AB)² + (BC)² = (AC)²}

\sf{ (11)² + (13)² = (18)² }

\sf{ 121 + 169 ≠  324 }

\sf{ 290 ≠ 324  }

<u>From </u><u>Above </u><u>we </u><u>can </u><u>conclude </u><u>that</u><u>, </u>

The sum of the squares of two small sides that is perpendicular height and base is not equal to the square of longest side that is Hypotenuse

\sf{\blue{ Hence,\: Your\: answer \:is \:false }}

5 0
2 years ago
Find the area of a square whose side in 7 inches (remeber A=b×h)
viva [34]
The area is 14. your welcome
7 0
3 years ago
Read 2 more answers
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