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daser333 [38]
3 years ago
6

500 in exponential form

Mathematics
1 answer:
allochka39001 [22]3 years ago
5 0

Answer:

5 x 102

Step-by-step explanation:

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I need help with this please and thank you!
Vlad [161]

Answer:  2x^2      -16t

               25t         -20

Step-by-step explanation:

Multiply the corresponding terms to find out which answer belongs in each box. For the top left box you would multiply  5t x 4t = 20t^2. For the top right multiply -4 x 4t = -16t. For bottom left box multiply 5t x 5 = 25t and for the bottom right box multiple -4 x 5 = -20.

6 0
3 years ago
If the point A at (5, 3) is rotated clockwise about the origin through 90o, what will be the coordinates of the new point?
evablogger [386]

The point (5, 3) is in first quadrant, then after the 90° clockwise the point will be in fourth quadrant and the coordinate will get swap. Then the coordinate will be (3, -5).

<h3>What is a transformation of a point?</h3>

A spatial transformation is each mapping of feature space to itself and it maintains some spatial correlation between figures.

If the point A at (5, 3) is rotated clockwise about the origin through 90°.

Then the coordinates of the new point will be

The point (5, 3) is in first quadrant, then after the 90° clockwise the point will be in fourth quadrant and the coordinate will get swap.

Then the coordinate will be (3, -5).

More about the transformation of a point link is given below.

brainly.com/question/27224339

#SPJ1

7 0
2 years ago
Multiplying with decimals
AnnyKZ [126]

Answer:

Dont know what your asking for

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Suppose the number of insect fragments in a chocolate bar follows a Poisson process with the expected number of fragments in a 2
leonid [27]

Answer:

a)The expected number of insect fragments in 1/4 of a 200-gram chocolate bar is 2.55

b)0.6004

c)19.607

Step-by-step explanation:

Let X denotes the number of fragments in 200 gm chocolate bar with expected number of fragments 10.2

X ~ Poisson(A) where \lambda = \frac{10.2}{200} = 0.051

a)We are supposed to find the expected number of insect fragments in 1/4 of a 200-gram chocolate bar

\frac{1}{4} \times 200 = 50

50 grams of bar contains expected fragments = \lambda x = 0.051 \times 50=2.55

So, the expected number of insect fragments in 1/4 of a 200-gram chocolate bar is 2.55

b) Now we are supposed to find the probability that you have to eat more than 10 grams of chocolate bar before ending your first fragment

Let X denotes the number of grams to be eaten before another fragment is detected.

P(X>10)= e^{-\lambda \times x}= e^{-0.051 \times 10}= e^{-0.51}=0.6004

c)The expected number of grams to be eaten before encountering the first fragments :

E(X)=\frac{1}{\lambda}=\frac{1}{0.051}=19.607 grams

7 0
3 years ago
How many solutions does 4x + 5 = 6(x + 3) - 20 - 2x have? Show your work. SOLUTION ​
liq [111]

Answer:

There are no solutions.

Step-by-step explanation:

4x+5=6x+18-20-2x (Distributive Property)

4x+2x-6x=18-20+5 (Combine Like Terms)

0=3 (Simplify)

That doesn't make any sense, so there are no solutions.

7 0
3 years ago
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