Can yall help meee please

Answers

Answer 1

Answer:

Answer:

$jupiters \: mass \: is \: approximatelty \: \n \boxed{{6 * 10}^(3) \: kg} \: times, \: \n more \: than \: mercurys \: mass$

Step-by-step explanation:

$............................................................... \n in \: other \: to \: get \: right \: ans \to \n you \: divide \: mass \: of \: jupiter \: by \: the \: mass \n \: of \: mercury : \: so \to \n ............................................................... \n jupiters \: mass = 1.898 { * 10}^(27) \: kg \n mercurys \: mass =3.3 * {10}^(23) \: kg \n their \: mass \: ratio \: is : \n = \frac{1.898 { * 10}^(27)}{3.3 * {10}^(23)} = 0.5751515152 * {10}^(4) = \n \boxed{5,751.515152} \n hence \: jupiters \: mass \: is \: approximatelty \: \n {6 * 10}^(3) \: times \: more \: than \: mercurys \: mass$

Answer 2

Answer: 6x103 because it explains that it’s 6 times but 10 in 3 times.

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Write a subtraction problem in which you have to rename a fraction and whosesolution is between 3 and 4

Answers

4 - 3 1/2 =1/2
is a possible answer

Find the real or imaginary solutions of the equation by factoring. x^4-8x^2=-16

Answers

ANSWER
$x=2$ or $x=-2$
EXPLANATION

The given equation is
${x}^(4) - 8 {x}^(2) = - 16.$

We set everything equal to zero to obtain,

${x}^(4) - 8 {x}^(2) + 16 = 0$

We rewrite the leading term to obtain,

${( {x}^(2) )}^(2) - 8( {x}^(2)) + 16 = 0$

This has now become a quadratic equation in
${x}^(2)$

We split the middle term to get,

${( {x}^(2) )}^(2) - 4 {x}^(2)- 4x^2+ 16 = 0$

We factor to get,

${x}^(2) (x^2-4)-4( {x}^(2)-4) = 0$

We factor further to get,

$( {x}^(2) -4)(x^2-4)= 0$

This gives

$( x-2)(x+2)(x-2)(x+2)= 0$

$( x-2)^2(x+2)^2= 0$

$x=2$ or $x=-2$

Line ZR bisects ∠QTS. Which is true? A. m∠QTR = m∠RTS B. m∠ZTS = m∠RTS C. m∠ZTQ = m∠QTR D. m∠STR < m∠QTR

Answers

Is the choice A: m∠QTR = m∠RTS]

Which problem is being modeled on the number line?

Answers

I don’t see a number line but I can help

Answer:

Step-by-step explanation:

Dianna has 7 quarters, 3 dimes, and 8 nickels in her purse. She reaches into her purse and randomly grabs two coins. What is the probability that Dianna grabs a dime first and then a nickel?

Answers

There are 18 coins total in her bag:

7+3+8=18coins

3 of the 18 coins are dimes, so the odds of first drawing a dime are 3/18

P(dime)=3/18

Since the nickel was taken out and not replaced, there are now 17 coins in her purse.

18-1=17coins

8 of the 17 remaining coins are nickels, so the odds of drawing a nickel are 8/17.

P(nickel)=8/17

Now we multiply the two probabilities by eachother to find out the odds of both occurring...

P(dime)*P(nickel)=

3/18*8/17=24/306=8/103=0.078=8%

The probability of grabbing a dime and then a nickel are 8/103 or 0.078 or an 8% chance, whichever way you'd prefer to write it.

Can someone please show how to solve this ln(5x-1)+ln(x-6)=ln6 ???

Answers

when you add ln, you multiply
ln((5x-1)(x-6)=ln6 do now FOIL
ln(5x^2-30x-x+6)=ln6
therefore
5x^2-31x+6=6 and
5x^2-31x=0 furthermore
x(5x-31)=0
one solution is 0, which we will reject
the other solution is 5x=31
x=31/5

ln(5x - 1) + ln(x - 6) = ln(6)
ln((5x - 1)(x - 6)) = ln(6)
      ln(5x(x - 6) - 1(x - 6) = ln(6)
ln(5x(x) - 5x(6) - 1(x) - 1(-6)) = ln(6)
ln(5x² - 30x - x + 6) = lm(6)
ln(5x² - 31x + 6) = ln(6)
   5x² - 31x + 6 = 6
5x² - 31x + 6 - 6 = 6 - 6
   5x² - 31x + 0 = 0
x = -(-31) +/- √((-31)² - 4(5)(0))
   2(5)
x = 31 +/- √(961 - 0)
   10
x = 31 +/- √(961)
10
x = 31 +/- 31
   10
x = 31 + 31 x = 31 - 31
10 10
x = 62    x = 0
10 10
x = 6²/₅ x = 0

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