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rjkz [21]
2 years ago
10

Pls help I’m giving away 10 points

Mathematics
1 answer:
notka56 [123]2 years ago
5 0

Answer:

20

Step-by-step explanation:

I can't see it but is y to the 1st power? If it is the answer is 20

PLZZ mark brainliest!!! Thank you :)

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Hi I hope you have a great day..... Can you please answer this question it would make me happy. :)
Blizzard [7]

Answer:

The answer is 11/20.

             

6 0
3 years ago
Please help me solve 8n+2-6n-13=9+2n-13
qwelly [4]
No solution since the 2n on both sides cross out
6 0
2 years ago
Read 2 more answers
An H0-scale model railroad engine is 9 inches long. The H0 scale is 87 feet to 1 foot. How long is a real engine?
Semmy [17]

The real engine is 65.25 feet long

<h3>How to determine the length?</h3>

The scale is given as:

Scale = 87 feet to 1 foot

The scale length is given as

Scale length = 9 inches

Convert inches to feet

Scale length = 9/12 feet

The actual length is then calculated as:

Actual = 87 * 9/12 feet

Evaluate

Actual = 65.25 feet

Hence, the real engine is 65.25 feet long

Read more about scale ratios at:

brainly.com/question/16192120

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7 0
1 year ago
Find the axis of symmetry and vertex of y=-2x^3+8​
sveta [45]

Answer:

Vertex:(0,8)     Axis of Symmetry: x=0

Step-by-step explanation:

7 0
3 years ago
A 0.45-mm-wide slit is illuminated by light of wavelength 590 nm. What is the width (in mm) of the central maximum on a screen 1
Hitman42 [59]

Answer:

w = \frac{2(590x10^{-9}) (1.5m)}{0.45x10^{-3}}=3.93 x10^{-3} m

And if we convert this into mm we got:

w= 3.93 x10^{-3} m *\frac{1000mm}{1 m}= 3.93 mm

Step-by-step explanation:

For this case we have the following info givenL

\lambda = 590 nm = 590 x10^{-9}m represent the wavelength

s = 0.45 mm *\frac{1m}{1000 mm}= 0.45x10^{-3} m represent the width of the single slit

L = 1.5 m represent the longitude

And we want to find the maximum width of the central maximum in mm, so we can use the following formula:

w s = 2 \lambda L

And if we solve for w the width we got:

w = \frac{2\lambda L}{s}

And replacing we got:

w = \frac{2(590x10^{-9}) (1.5m)}{0.45x10^{-3}}=3.93 x10^{-3} m

And if we convert this into mm we got:

w= 3.93 x10^{-3} m *\frac{1000mm}{1 m}= 3.93 mm

6 0
3 years ago
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