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babunello [35]
3 years ago
10

Solve the equation.

Mathematics
1 answer:
morpeh [17]3 years ago
4 0

Answer:x=0

Step-by-step explanation:

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if j is the number of integers between 1 and 500 that are divisible by 9 and k is the number of integers between 1 and 500 that
Morgarella [4.7K]

Answer:

  126

Step-by-step explanation:

The number of numbers divisible by 9 is ...

  j = floor(500/9) = 55

The number of numbers divisible by 7 is ...

  k = floor(500/7) = 71

Then the total (j+k) is ...

  j +k = 55 +71 = 126

3 0
4 years ago
How many 6-digit binary strings have the property that their second and fourth digits are 1s?
Tresset [83]
Fixing two of the bits leaves four remaining positions, each with two possibilities (0 or 1), so there are 2^4=16 possible strings.
3 0
3 years ago
25. The width of a rectangle is (x+1) and the length is (X-6). What is the length and width of the rectangle if the area is
Elis [28]

Answer:

x=−4 or x=9

Step-by-step explanation:

1- As we know the formula for Area of rectangle = length x width

2- Multiply given width = (x+1) with length = (x-6)

3-Equation will be (x+1)(x-6) = 30

4- Solving x^2−5 x−6  = 30

5- x^2−5 x -36 = 0

6-Factor left side of equation.

(x+4)(x−9)=0

Set factors equal to 0.

x+4=0 or x−9=0

6 0
3 years ago
Read 2 more answers
Jane is is charge of making a banner for the basketball game this Saturday. She measures how long the banner is before painting
RSB [31]

Answer:

The measurement which Jane finds to be 10 meters is the length of the banner.

Step-by-step explanation:

The measurement of 10 meters which Jane found after measuring how long the banner is before painting is the LENGTH of the banner.

This is clear from the unit of what she finds (meters). It only indicates the measurement of one part of the banner, even though a banner has two parts, the length and width.

It is possible to find the AREA, or PERIMETER, or LENGTH.

But what she finds is the LENGTH of the banner. If it was Area or Perimeter, the unit would have been square meters.

7 0
3 years ago
Prove that if an integer is a perfect square and a perfect cube, simultaneously, then it is either of the form 7k or 7k+1
Bogdan [553]
If an integer is both a square and a cube, it can be of the form:
<span>(<span>a3</span><span>)^2</span></span>
Now,
since a cube can be of the form 7k or 7k+-1(thanks to FoolForMath),
 we write
<span><span>a^3</span>=7k</span>
and get the no to be
49k^2
, which is in the form of 7 times something
<span>49<span>k^2</span>=7×(7<span>k^2</span>)</span>
Now put
<span><span>a^3</span>=7k+−1</span> Square it
and you'll get a number in the form of (7times something +1)
5 0
3 years ago
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