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4 years ago

Use some of the digits 1,3,5,7,9 only once to write 5 numbers less than 2000

Mathematics

1 answer:

Yuri [45]4 years ago

3 0

Answer:

1357, 1359, 1579, 1573, 1375, 1379, 1957, 1953

Step-by-step explanation:

Since this number needs to be less than 2000, the first number has to be 1. You can randomly pick and choose any of the numbers after that.

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What is the value of a in the equation a = 2 + 3a + 10?

xxTIMURxx [149]

Answer:

a= -6

Step-by-step explanation:

Given

a=2+3a+10

In order to find the value of a, we have to isolate a on a single side of the equation

So,

subtracting 3a from both sides

a-3a = 2+10+3a-3a

=>a-3a=12

=> -2a = 12

Dividing both sides by -2

=> -2a/-2 = 12/-2

=> a = -6

The value of a in the given equation is -6 ..

3 0

3 years ago

Write the slope intercept form that satisfies the following conditions

Olin [163]

$y = 3x - 13$

Step-by-step explanation:

The line will have the same slope as the given line, which is m = 3. Therefore, the equation of the line in its slope-intercept form is

$y = 3x + b$

Next we need to solve for b by plugging in the coordinates of the given point:

$(-4) = 3(3) + b \Rightarrow b = -13$

So the equation of the line is

$y = 3x - 13$

7 0

2 years ago

What is the first step in solving

netineya [11]

Answer:

Step-by-step explanation:

4 0

4 years ago

Find the 10th and 21st terms of -3,1,5,9

r-ruslan [8.4K]

Answer:

33, 77

Step-by-step explanation:

-3. 1 5 9 13 17 21 25 29 33 37 41 45 49 53 57 61 65 69 73 77

4 0

3 years ago

Read 2 more answers

The circumference of the ellipse approximate. Which equation is the result of solving the formula of the circumference for b?

Serhud [2]

Answer:

$b = \sqrt{\frac{C^{2} }{2(\pi )^{2} } - a^{2}}$

Step-by-step explanation:

Given - The circumference of the ellipse approximated by $C = 2\pi \sqrt{\frac{a^{2} + b^{2} }{2} }$ where 2a and 2b are the lengths of 2 the axes of the ellipse.

To find - Which equation is the result of solving the formula of the circumference for b ?

Solution -

$C = 2\pi \sqrt{\frac{a^{2} + b^{2} }{2} }\\\frac{C}{2\pi } = \sqrt{\frac{a^{2} + b^{2} }{2} }$

Squaring Both sides, we get

$[\frac{C}{2\pi }]^{2} = [\sqrt{\frac{a^{2} + b^{2} }{2} }]^{2} \\\frac{C^{2} }{(2\pi)^{2} } = {\frac{a^{2} + b^{2} }{2} }\\2\frac{C^{2} }{4(\pi)^{2} } = {{a^{2} + b^{2} }$

$\frac{C^{2} }{2(\pi )^{2} } = a^{2} + b^{2} \\\frac{C^{2} }{2(\pi )^{2} } - a^{2} = b^{2} \\\sqrt{\frac{C^{2} }{2(\pi )^{2} } - a^{2}} = b$

∴ we get

$b = \sqrt{\frac{C^{2} }{2(\pi )^{2} } - a^{2}}$

8 0

3 years ago