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3 years ago

X/9 = 9/27 please solve this for me

Mathematics

2 answers:

mars1129 [50]3 years ago

6 0

Answer:

x=3

Step-by-step explanation:

This is a proportional equation:

Given: x/9=9/27

Cross-multiply: 27x=81

Divide: x=3

So x=3

Ivenika [448]3 years ago

6 0

Answer:

x=3

Step-by-step explanation:

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I know number 13 is right, how do I do 11 and 12?

snow_tiger [21]

11. 43 ; 137
12. 12 ; 102

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3 years ago

In un trapezio rettangolo la base minore, il lato obliquo e l'altezza misurano rispettivamente 60 cm. 95 cm è 76 cm. Calcola il

KengaRu [80]

Answer:

$2p=348 cm\: S=6726 cm^{2}$

Step-by-step explanation:

Ciao, come stai?

1) Per prima cosa, dobbiamo trovare la misura della base più grande. Scomponendo la figura possiamo visualizzare un triangolo e un quadrato. Ci sono somiglianze con gli angoli. Quindi è un triangolo rettangolo. Applichiamo il teorema di Pitagora:

$a^2=b^2+c^2\\95^2=b^2+76^2\\57=b$

2) Perimetro:

$2p=60+57+76+60+95\\2p=348 cm$

3) L' area

$\frac{(B+b)h}{2} =\frac{(117+60)76}{2} =6726 \:cm^{2}$

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3 years ago

Truth or Dare? Have a nice day/night <3

denpristay [2]

Answer:

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Step-by-step explanation:

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3 years ago

Read 2 more answers

What is the greatest common factor of 44c^5, 22c^3 and 11c^4

Alexxandr [17]

$\text{ GCF of } 44c^5, 22c^3 , 11c^4 \text{ is } 11c^3$

Solution:

Given that, we have to find the greatest common factor of $44c^5, 22c^3 , 11c^4$

The greatest number that is a factor of two (or more) other numbers.

When we find all the factors of two or more numbers, and some factors are the same ("common"), then the largest of those common factors is the Greatest Common Factor

$\underline{\text{ GCF of } 44c^5, 22c^3 , 11c^4 :}$

First find the GCF of numbers and then find the GCF of variables and then multiply them together

Step 1: GCF of 44, 22 and 11

The factors of 11 are: 1, 11

The factors of 22 are: 1, 2, 11, 22

The factors of 44 are: 1, 2, 4, 11, 22, 44

11 is the greatest factor that is common in above three list

Then the greatest common factor is 11

$\text{ Step 2: GCF of }c^5, c^3 , c^4$

$c^5 = c^3 \times c^2\\\\c^3 = c^2 \times c^1 \text{ or } c^3\\\\c^4 = c^3 \times c^1$

Thus the greatest common factor is $c^3$

Step 3: Multiply the GCF of 44, 22, 11 and GCF of $c^5, c^3 , c^4$

$\text{ GCF of } 44c^5, 22c^3 , 11c^4 = 11 \times c^3 = 11c^3$

Thus $\text{ GCF of } 44c^5, 22c^3 , 11c^4 \text{ is } 11c^3$

8 0

3 years ago

Someone pls do these 3 for me, its due in 20 minutes and my head hurts so bad right now :(

Drupady [299]

Answer:

1. Chloe has more money.

2. Dylan does not have enough money to buy 3 bags of candy.

3. Mikayla could get 4 pennies, 2 nickels, 2 dimes, and 2 quarters back in change.

Step-by-step explanation:

Penny = 1 cent

Nickel = 5 cents

Dime = 10 cents

Quarter = 25 cents

1. Chloe has 2 quarters(50 cents) and 3 dimes(30 cents). 50 + 30 = 80 cents

Haru has 5 nickels(25 cents) and 4 dimes(40 cents). 25 + 40 = 65 cents

2. If each bag costs $350(oddly enough). Dylan would need $1,050 to buy 3 bags of candy. Instead he only has $10. This means he would need $1,040 more dollars.

3. $5 - 4.16 = 84 cents in return

quarter(25) + quarter(25) + dime(10) + dime(10) + nickel(5) + nickel(5) + penny(1) + penny(1) + penny(1) + penny(1) = 84 cents

3 0

2 years ago