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2 years ago

What the common multiple of 10

Mathematics

1 answer:

Viktor [21]2 years ago

5 0

Answer:

20 30 40 40 50 60 70 80 90 100

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the sum of two numbers is 63. one ninth of the first number plus one sixth of the second number is 21. find the numbers

PSYCHO15rus [73]

Answer: x= -189, y =252

Step-by-step explanation:

Let the first number be x and second number be y

x + y = 63

x = 63-y

now

1/9 of x + 1/6 of y = 21

x/9 + y/6 = 21

substituting x's value from equation i

(63-y)/9 + y/6 = 21

(378-6y+9y)/54 = 21

378+3y = 1134

3y = 1134-378

so, 3y = 756

so, y = 756/3

so, y = 252

now

x = 63-252

so, x = -189

6 0

2 years ago

I'll go on a date with anyone who can answer this and help me...?

vesna_86 [32]

Answer:

Chill is this a multiple choice?

7 0

3 years ago

If tom had 100g rice how would u convert it to kg

notka56 [123]

répondre:

bonjour

si tom avait 100 g de riz il va convertir 100g en kg

Étape par Étape explication:

= 0,1 kg

bonne journée

3 0

2 years ago

Covert each into algebraic equations

marshall27 [118]

Answer:

1) 2x=4x-8

2) 4(x+2)=20

3) 2x-9=4(x+5)

3 0

2 years ago

Answer correctly please

Paul [167]

Answer:

$\$272,49$

Step-by-step explanation:

7. $\displaystyle /text{The answer makes sense because since the depreciation rate is 15%, we know that we need to use the "exponential decay" formula.}$

6. $\displaystyle /text{After a depreciation rate of 15% for the past 8 years, the stock is now worth approximately $272,49.}$

5. $\displaystyle 1000[0,85]^8 = 272,490525 ≈ \$272,49$

4. $\displaystyle 1000 = a \\ -15\% + 100\% = 1 - r; 85\% = 1 - r \\ 8\:years = time\:[t]$

3. $\displaystyle /text{We need to use the "Exponential Decay" formula} - f(t) = a[1 - r]^t, where a > 0$

2. $\displaystyle /text{How much is the stock worth after a depreciation rate of 15% per year?}$

1. $\displaystyle /text{initial amount: $1000, a depreciation rate of 15%, and a time period of 8 years}$

I am joyous to assist you anytime.

5 0

3 years ago