All categories

3 years ago

It rained 6.14 inches on Saturday and 4.71 inches on Sunday. How much did it rain on

Mathematics

1 answer:

soldier1979 [14.2K]3 years ago

6 0

Answer:

10.65 inches

Step-by-step explanation:

You simply add Saturday's and Sunday's total together

6.14+4.71=10.85

You might be interested in

What is the value of the expression? (-3)(-4)(-5)

nekit [7.7K]

Answer:

-60

Step-by-step explanation:

-3 x -4 x -5

12 x -5

-60

negative x negative x negative= negative

4 0

3 years ago

Greg has enough framing to enclose a rectangular photograph with a perimeter of 160 centimeters. If the width of his photograph

kotegsom [21]

Answer:

32cm

Step-by-step explanation:

Let the width be w. Now we know that the width is 16cm less than the length, this means that the length will thus have a total length of w + 16

Now, we know that the perimeter of a rectangle is 2(l + b)

This means 2(w + w + 16) = 160

4w + 32 = 160

4w = 128

w = 128/4 = 32cm

8 0

4 years ago

In my town, gas prices are always listed to the thousandths place. Since the smallest coin we have is the penny, we have to roun

AlladinOne [14]

Answer:

$8.16

Step-by-step explanation:

6 0

3 years ago

Determine all prime numbers a, b and c for which the expression a ^ 2 + b ^ 2 + c ^ 2 - 1 is a perfect square .

kogti [31]

Answer:

The family of all prime numbers such that $a^{2} + b^{2} + c^{2} -1$ is a perfect square is represented by the following solution:

$a$ is an arbitrary prime number. (1)

$b = \sqrt{1 + 2\cdot a \cdot c}$ (2)

$c$ is another arbitrary prime number. (3)

Step-by-step explanation:

From Algebra we know that a second order polynomial is a perfect square if and only if $(x+y)^{2} = x^{2} + 2\cdot x\cdot y + y^{2}$ . From statement, we must fulfill the following identity:

$a^{2} + b^{2} + c^{2} - 1 = x^{2} + 2\cdot x\cdot y + y^{2}$

By Associative and Commutative properties, we can reorganize the expression as follows:

$a^{2} + (b^{2}-1) + c^{2} = x^{2} + 2\cdot x \cdot y + y^{2}$ (1)

Then, we have the following system of equations:

$x = a$ (2)

$(b^{2}-1) = 2\cdot x\cdot y$ (3)

$y = c$ (4)

By (2) and (4) in (3), we have the following expression:

$(b^{2} - 1) = 2\cdot a \cdot c$

$b^{2} = 1 + 2\cdot a \cdot c$

$b = \sqrt{1 + 2\cdot a\cdot c}$

From Number Theory, we remember that a number is prime if and only if is divisible both by 1 and by itself. Then, $a, b, c > 1$ . If $a$ , $b$ and $c$ are prime numbers, then $2\cdot a\cdot c$ must be an even composite number, which means that $a$ and $c$ can be either both odd numbers or a even number and a odd number. In the family of prime numbers, the only even number is 2.