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3 years ago

Can you help me please

Mathematics

1 answer:

katrin2010 [14]3 years ago

3 0

Answer:

1:) x=6

2:)x= -10

3:) x=8

4:)x=-4

Step-by-step explanation:

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Solve for x in simplest form 15 = 1/3(x+9)

Lorico [155]

Heya!

Your question states: Solve for x in simplest form 15 = 1/3(x+9)

Answer: X = 36

<em>Numerical Explanation:</em>

1. Step by step

I. 15 = 1/3(x+9)

2. Simplify both sides of the equation:

II. 15 = 1/3x + 3

3. Flip the equation:

III. 1/3x + 3 = 15

4. Subtract 3 from both sides:

IV. 1/3 x +3 = 15

-3 -3

5. Multiply both sides by 3:

V. 3 * (1/3 x) = (3) * (12)

6. Final answer:

VI. x = 36

I Hoped I Helped!

~KINGJUPITER

6 0

3 years ago

Read 2 more answers

What measurement is equal to 20 liters in milliliters?

kakasveta [241]

Litersx1000=mls
20x1000=20000mls

5 0

2 years ago

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Cressida decided to save a part of her income in a savings account that compounds interest every four months. Her account balanc

Bogdan [553]

Answer:

Step-by-step explanation:

account balance on compound interest = principal amount (1+rate/100)^time

here there are t years so toat months are 12t and ad interest is after each 4 months so time is 12t/4 =3t

now total balance = 3545(1.045)^3t

here 3545 is principal amount , rate is( 1.045-1)×100 =4.5% per 4 months and 3t is time so clearly 3t or 3 is a growth factor hence option A

8 0

3 years ago

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A survey of teens suggested that 33% can name at least one professional baseball player, and 90% of those teens can also name at

loris [4]

Answer:

$d.\ 0.33 + 0.64 – (0.90)(0.33) = 0.673$

Step-by-step explanation:

Given

$P(Baseball) = 33\%$

$P(Baseball\ n\ Football) = 90\%$

$P(Football) = 64\%$

Required

$P(Baseball\ or\ Football)$

This is calculated as:

$P(Baseball\ or\ Football) =P(Baseball) + P(Football) -P(Baseball\ n\ Football)$

$P(Baseball\ or\ Football) = 0.33 + 0.64 - (0.90)(0.33)$

$P(Baseball\ or\ Football) = 0.33 + 0.64 - 0.297$

$P(Baseball\ or\ Football) = 0.673$

3 0

2 years ago

The rate of change of the number of squirrels S(t) that live on the Lehman College campus is directly proportional to 30 − S(t),

pishuonlain [190]

Answer:

S(3)=22

Step-by-step explanation:

The rate of change of the number of squirrels S(t) that live on the Lehman College campus is directly proportional to 30 − S(t).

$\dfrac{dS}{dt}=k(30-S(t))\\ \dfrac{dS}{dt}+kS(t)=30k\\$The integrating factor: e^{\int k dt}=e^{kt}\\$Multiply all through by the integrating factor\\ \dfrac{dS}{dt}e^{kt}+kS(t)e^{kt}=30ke^{kt}$

$(Se^{kt})'=30ke^{kt} dt\\$Integrate both sides\\ Se^{kt}=\dfrac{30ke^{kt}}{k}+C$ (C a constant of integration)\\Se^{kt}=30e^{kt}+C\\$Divide both sides by e^{kt}\\S(t)=30+Ce^{-kt}$

When t=0, S(t)=15

$15=30+Ce^{-k*0}\\C=15-30\\C=-15$

When t = 2, S(t)=20

$20=30-15e^{-2k}\\20-30=-15e^{-2k}\\-10=-15e^{-2k}\\e^{-2k}=\dfrac23\\$Take the natural log of both sides$\\-2k=\ln \dfrac23\\k=-\dfrac{\ln(2/3)}{2}$

Therefore:

$S(t)=30-15e^{\frac{\ln(2/3)}{2}t}\\$When t=3$\\S(t)=30-15e^{\frac{\ln(2/3)}{2} \times 3}\\S(3)=21.8 \approx 22$

8 0

2 years ago