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3 years ago

Express in decimal numerals.- nine and two hundred thirty-five thousands

Mathematics

2 answers:

Vinil7 [7]3 years ago

8 0

902.350 that's your answer

AleksandrR [38]3 years ago

5 0

Well the question is not so clear put I will give it a try
9,235,000
I am pretty sure that's how you write it

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For all values of x,<br> x² + 6x - 2 = (x+p)² + q<br> Find the value of p and the value of q.

Vlad [161]

Step-by-step explanation:

${x}^{2} + 6x - 2 = {x}^{2} + 2px + {p}^{2} + q$

$2p = 6$

${p}^{2} + q = - 2$

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${3}^{2} + q = - 2$

$q = - 11$

p=3, q=-11

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2 years ago

What is the solution to the inequality -4x < 8? x < -24 x > -24 x < -2 x > -2

liberstina [14]

Answer:

x > -2

Step-by-step explanation:

Solve the inequality -4x < 8 using inverse operations. If you divide or multiply by a negative, flip the inequality sign.

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3 years ago

Marley passes the library on her way to school. The distance from marleys house to the library is 3/8 mile. The distance from ma

irina [24]

0/7 just subtract them pretty sure

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3 years ago

Three coins are flipped.

Debora [2.8K]

Answer:

3/8

Step-by-step explanation:

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3 0

3 years ago

Read 2 more answers

You have a chess board as pictured, with squares numbered 1 through 64. You also have a huge change jar with an unlimited number

ANEK [815]

Answer:

(a) 512 dimes

(b) $D(n) = 2^{n-1}$

(d) 9.22337 x 10¹² km

(e) roughly 61,489.13 times longer than the distance from the earth to the sun.

Step-by-step explanation:

b. Starting with 1 dime in the first square, if the number of dimes per each square is the double of the previous square, the general equation for the number of dimes in each square 'n' can be found by:

$D(1) = 1\\D(2) = 2^1\\D(3) = 2^2\\D(4) = 2^3\\D(n) = 2^{n-1}$

a. On the 10th square:

$D(10) = 2^{10-1}\\D(10) = 512\ dimes$

c. On the 64th square:

$D(64) = 2^{64-1}\\D(10) = 9.22337 *10^{18}\ dimes$

d. If a dime is 1 mm thick, the 64th pile will be:

$h= 9.22337*10^{18}*0.0001\ m\\h= 9.22337*10^{15}\ m\\h= 9.22337*10^{12}\ km$

e. The ratio between the height of the last pile (h) to the distance from the earth to the sun is:

$R=\frac{9.22337*10^{12}\ km}{150,000,000\ km}\\ R=61,489.13$

The height of the last pile is roughly 61,489.13 times longer than the distance from the earth to the sun.

5 0

3 years ago