All categories

3 years ago

What is the remainder of 28 ÷ 13

Mathematics

2 answers:

Sonbull [250]3 years ago

8 0

Answer:

Step-by-step explanation:

28 ÷ 13 → 2 (quotient)

13 * 2 = 26

28 - 26 = 2 (remainder)

Arte-miy333 [17]3 years ago

4 0

13 goes into 28 evenly two times - 13 x 2 = 26

What's left is 28 - 26 = 2

The remainder of 28 / 13 is 2.

Hope this helps!! :)

You might be interested in

Un triángulo rectángulo tiene un área de 84 pies2 y una hipotenusa de 25 pies de largo. ¿Cuáles son las longitudes de sus otros

jolli1 [7]

Answer:

7 feet and 24 feet

Step-by-step explanation:

In the right triangle, the hypotenuse is 25 feet long. The area of this triangle is 84 square feet.

Let x feet and y feet be the lengths of triangle's legs.

By the Pythagorean theorem,

$x^2+y^2=25^2\\ \\x^2+y^2=625$

The area of the right triangle is half the product of its legs, thus

$84=\dfrac{1}{2}xy\\ \\xy=168$

Solve the system of two equations:

$\left\{\begin{array}{l}x^2+y^2=625\\ \\xy=168\end{array}\right.$

From the second equation:

$x=\dfrac{168}{y}$

Substitute it into the first equation:

$\left(\dfrac{168}{y}\right)^2+y^2=625\\ \\168^2+y^4=625y^2\\ \\y^4-625y^2+168^2=0\\ \\D=(-625)^2-4\cdot 168^2=(625-2\cdot 168)(625+2\cdot 168)=289\cdot 961=17^2\cdot 31\\ \\\sqrt{D}=17\cdot 31=527\\ \\y^2_{1,2}=\dfrac{-(-625)\pm 527}{2}=49,\ 576\\ \\y_1=7,\ y_2=-7,\ y_3=24,\ y_4=-24$

The length of the leg cannot be negative, so

$y_1=7\Rightarrow x_1=\dfrac{168}{7}=24\\ \\y_2=24\Rightarrow x_2=\dfrac{168}{24}=7$

6 0

3 years ago

HELP HELP PLEASE omgggg

Vlad [161]

Answer:

Step-by-step explanation:

There are 2 rays GF and GA

The line is AC

The Segment is CE. You could argue that GC is also a segment. I don't think it is the intended answer. As an answer, CE is much safer.

3 0

3 years ago

Set up the integral that uses the method of disks/washers to find the volume V of the solid obtained by rotating the region boun

Blababa [14]

Answer:

$a)\: \frac{3}{2} \pi$

$b) \:\frac{47}{5} \pi$

Step-by-step explanation:

$y=3\sqrt x,\:\:y=3,\:\:x=0$

a) about the line y = 3

$3\sqrt x=3$ ⇒ $x = 1$ is the intersection point

So,

$V=\int\limits^1_0\pi(3-3\sqrt x)^2\:dx=\pi\int\limits^1_0(9-18\sqrt x+9x)\:dx=\\\\=\pi(9x-12x^{3/2}+9/2x^2)|^1_0=\pi(9-12+9/2)=\frac{3}{2} \pi$

b) about the line x = 5

$y=3\sqrt x$ ⇒ $x=y^2/9$

So,

$V = \int\limits^3_0\pi([5-0]^2-[5-y^2/9]^2)\:dy=\pi\int\limits^3_0(25-25+10y^2/9-y^4/81)\:dy=\\\\=\pi(10y^3/27-y^5/405)|^3_0=\pi(10-3/5)=\frac{47}{5} \pi$