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9966 [12]
4 years ago
14

When you multiply my number by 6 and add 7 you get 25 what is my number

Mathematics
1 answer:
pav-90 [236]4 years ago
8 0
To solve this you do 25-7=18 and do 18÷6=3, so my number is 3. To check if this is correct you do 3×6=18 and 18+7=25
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Both circle A and circle B have a central angle measuring 50°. The area of circle A's sector is 36π cm2, and the area of circle
slega [8]

Answer:

A) 3/4

Step-by-step explanation:

Given: Both circle A and circle B have a central angle measuring 50°.

           The area of circle A's sector is 36π cm2.

            The area of circle B's sector is 64π cm2.

We know, area of the circle= \pi r^{2}

lets assume the radius of circle A be "r_1" and radius of circle B be "r_2"

As given, Area of circle A and B´s sector is 36π and 64π repectively.

Now, writing ratio of area of circle A and B, to find the ratio of radius.

⇒\frac{\pi r_1^{2}  }{\pi r_2^{2} }  = \frac{36\pi }{64\pi }

Cancelling out the common factor

⇒ \frac{r_1^{2}  }{r_2^{2} }  = \frac{36 }{64}

⇒ (\frac{r_1  }{r_2} )^{2} = \frac{36 }{64}

Taking square on both side.

Remember; √a²= a

⇒ (\frac{r_1  }{r_2} ) =\sqrt{ \frac{36 }{64}}

⇒ (\frac{r_1  }{r_2} ) = \frac{6}{8}

⇒\frac{r_1  }{r_2}  = \frac{3}{4}

Hence, ratio of the radius of circle A to the radius of circle B is 3:4 or 3/4.

5 0
3 years ago
han want to buy a $30 ticket to a game, but the pre-oder tickets are sold out, he knowns there will be more tickets sold the day
Kay [80]

Answer: \$90

Step-by-step explanation:

Given

The price of the ticket prior to match day is \$30

It is stated that there is 200\% the markup on match day i.e.

The selling price of the ticket is

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6 0
3 years ago
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tekilochka [14]

Answer with Step-by-step explanation:

We are given that A, B and C are subsets of universal set U.

We have to prove that

A\cap (B-C)=(A\cap B)-(A\cap C)

Proof:

Let x\in A\cap (B-C)

Then x\in A and x\in(B-C)

When x\in ( B-C)then x\in B but x\notin C

Therefore, x\in( A\cap B) but x\notin (A\cap C)

Hence, it is true.

Conversely , Let x\in(A\cap B) but x\notin(A\cap C)

Then x\in A and x\in B

When x\notin ( A\cap C) then x\notin C

Therefor,x\in A\cap (B-C)

Hence, the statement is true.

5 0
3 years ago
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Bess [88]

Answer:

3.38 cm

Step-by-step explanation:

will be

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3 years ago
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The way you wrote the question is unable to understand
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