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muminat
3 years ago
10

Nathaniel started collecting rocks. He started his collection with two rocks on the first day, then four rocks on the second day

, then
eight rocks on the third day, and then sixteen rocks on the fourth day.
Assuming this pattern continues, what will be the total size of Nathaniel's rock collection on the 10th day?
Mathematics
1 answer:
Marizza181 [45]3 years ago
5 0

Answer:

the answer is 1024 I just got it right trust me

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If yall can answer this correctly, ill give you extra pointssss!
horrorfan [7]

Answer:

23

Step-by-step explanation:

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3 years ago
Click on the value for p that makes this<br> inequality true.<br><br><br><br> Pls I need this
weqwewe [10]

p = unknown number

p ≥ -12     ["p" is any number greater than or equal to -12]

The only value that can make this true is 0, because all the other numbers are less than -12.

0 ≥ -12   [0 is greater than or equal to -12]

5 0
3 years ago
Read 2 more answers
What is the approximate percent decrease when a fish count goes down from 850 to 500?
Fudgin [204]
So, the count went down from 850 to 500, or 350 units.

if we take the 850 to be the 100%, what is 350 in percentage off of it?

\bf \begin{array}{ccllll}&#10;amount&\%\\&#10;\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\&#10;850&100\\&#10;350&x&#10;\end{array} \implies \cfrac{850}{350}=\cfrac{100}{x}

solve for "x".
5 0
3 years ago
Look at the residual plot below. Does it indicate a GOOD fit or a BAD fit for a linear model?
Triss [41]

Answer:

-2.0098787

Step-by-step explanation:

1+1=2

-2=-2

27531651721863518/-2=

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7 0
4 years ago
An AISI 1040 cold-drawn steel tube has an OD 5 50 mm and wall thickness 6 mm. What maximum external pressure can this tube withs
lukranit [14]

Answer:

82.79MPa

Step-by-step explanation:

Minimum yield strength for AISI 1040 cold drawn steel as obtained from literature, Sᵧ = 490 MPa

Given, outer radius, r₀ = 25mm = 0.025m, thickness = 6mm = 0.006m, internal radius, rᵢ = 19mm = 0.019m,

Largest allowable stress = 0.8(-490) = -392 MPa (minus sign because of compressive nature of the stress)

The tangential stress, σₜ = - ((r₀²p₀)/(r₀² - rᵢ²))(1 + (rᵢ²/r²))

But the maximum tangential stress will occur on the internal diameter of the tube, where r = rᵢ

σₜₘₐₓ = -2(r₀²p₀)/(r₀² - rᵢ²)

p₀ = - σₜₘₐₓ(r₀² - rᵢ²)/2(r₀²) = -392(0.025² - 0.019²)/2(0.025²) = 82.79 MPa.

Hope this helps!!

8 0
4 years ago
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