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FrozenT [24]
2 years ago
9

PLS HELPP THANKK UUU

Mathematics
1 answer:
pantera1 [17]2 years ago
4 0

Answer: Heyaa!! :) ^^

Your Answer Is... y=−6−2x

Step-by-step explanation:

Move all terms that don't contain y to the right side and solve.

Add 2x to both sides of the equation.

  • −y= 6+2x

Divide each term in −y=6+2x by −1 and simplify.

<h3>^ y=−6−2x ^</h3>

Hopefully this helps <em>you !</em>

<em />

<em> Matthew</em>

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Which exponential expression is equal to 2−5·28
gizmo_the_mogwai [7]

Answer:

Impossible

Step-by-step explanation:

Solve the answer 2-5*28 = -138

This would have no answer since an exponential expression can't equal an negative unless it's a negative answer, but in this case, there is no answer that would get you -138.

3 0
3 years ago
Sally is analyzing a circle, y2 + x2 = 9, and a linear function g(x). Will they intersect?
Mamont248 [21]
This ones kinda hard I'm not really sure, but looking at the table, when f(x) = 1, g(x) = 1. So therefore it is yes, and Im guessing you know the negative and positive x coordinates/zero thing, so I think you should be correct. Sorry if this is wrong, not too sure, but hopefully it gives you a better idea.
4 0
3 years ago
Read 2 more answers
Keith has 39 books in his library he brought several books at a yard sale over the weekend he now has 84 in his library how many
garri49 [273]

Subtract original amount from new amount:

84 - 39 = 45

He bought 45 books.

3 0
3 years ago
Write the equation of the line passing through the points (2,-5) and (7,-3).
Grace [21]

Answer:

5y·2x=21

<em>\frac{y - y1}{y2 - y1}  =  \frac{x - x1}{x2 - x1}</em>

Step-by-step explanation:

using the formula above,you will find the equation 5y–2x=21

7 0
3 years ago
Complete the table for the radioactive iotope. (Round your anwer to 2 decimal place. Iotope : 239 Pu
Marizza181 [45]

If the radio active isotope has the half life of 24100 years, then the initial quantity is 2.16 grams

The half life in years = 24100

Consider the quantity of the radio active isotope remaining

y = Ce^{kt}

When t = 1000 the y = 1.2

y = C/2 when t = 1599

Substitute the values in the equation

C/2 = Ce^{24100k}

Cancel the C in both side

1/2 = e^{24100k}

Here we have to apply ln to eliminate the e terms

ln (1/2) = 24100k

k = ln(1/2) / 24100

k = -2.87× 10^-5

To find the initial value we have to substitute the value of k and y in the equation

1.2 = Ce^{1000 ×  -2.87× 10^-5}

C = 1.2 / e^(-0.0287)

C = 2.16 gram

Hence, the initial quantity of the radioactive isotope is 2.16 gram

Learn more about half life here

brainly.com/question/4318844

#SPJ4

7 0
1 year ago
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