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3 years ago

Five less than Greg's age is less than 23

Mathematics

1 answer:

igomit [66]3 years ago

6 0

Answer:

Greg is 28.

Step-by-step explanation:

Greg's age is y.

y - 5 = 23

y - 5 + 5 = 23 + 5

y = 28

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PLEASE TELL ME THE ANSWER IM CONFUSED

emmainna [20.7K]

Answer:

$A=150.72$ $m^2$

Step-by-step explanation:

Using the formula for the surface area of a cylinder

$A=2\pi rh+2\pi r^2$

As we know that r=3, h=5, and $\pi$ =3.14 we can substitute in these values into the equation and then solve it

$A=2(3.14)(3)(5)+2(3.14)(3)^2\\\\A=94.20+56.52\\\\A=150.72$

3 0

3 years ago

Answer quick<br><br> Calculate the slope of the line going through A(-4,3) and B(0,6)

NeX [460]

Answer:

Step-by-step explanation:

$\boxed{slope = \frac{y1 - y2}{x1 - x2} }$

☆ (x₁, y₁) is the first coordinate and (x₂, y₂) is the second coordinate

Slope of line

$= \frac{6 - 3}{0 - ( - 4)}$

$=\frac{3}{4}$

3 0

3 years ago

Ted has $3.90. He has only dimes and quarters. He has 27 coins altogether. How many of each does he have

mestny [16]

Answer: 8 quarters 19 dimes

Step-by-step explanation:

we already know 8+19=27, if 4 quarters equals a dollar then you use 4 quarters each to create 2 dollars.

then, use 10 dimes to create another dollar summing it up to $3. now all we have to do is make 90 cents, and we've already used 18 coins.

if we used another 9 dimes, that would equal 90 cents.

in conclusion, 4+4= 8

8+10= 18

18+9= 27, and having $3.90

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.