All categories

4 years ago

What is the length of the base of a right triangle with an area of 20 m^2 and a height of 4 m?

Mathematics

1 answer:

Mariana [72]4 years ago

4 0

Answer:

The Length of the right triangle is B, 10

Step-by-step explanation:

I took the test ;) Hope this helps!

You might be interested in

Find the area of a trapezium shaped table whose parallel sides are 6 metre and 8 metre and the distance between the parallel sid

katovenus [111]

Length of parallel sides=8 cm, 9 cm
Distance between parallel sides=6 cm
Area =
2
1

(distance between parallel sides)× (sum of parallel sides)
=
2
1

×6×(8+9)
=3×17=51

hope it helps

5 0

3 years ago

Pls join this one pls no pervs or people over 13 PLS OMG

rjkz [21]

Answer:

Bet

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Every integer is a______

Anit [1.1K]

Answer:

the answer is b. rational

5 0

4 years ago

Equivalent fraction of 7/21

Radda [10]

Answer:

1/3

Step-by-step explanation:

7 x 3 = 21

21 / 3 = 7

3 0

3 years ago

Read 2 more answers

Let $f(x) = x^2$ and $g(x) = \sqrt{x}$. Find the area bounded by $f(x)$ and $g(x).$

Anna [14]

Answer:

$\large\boxed{1\dfrac{1}{3}\ u^2}$

Step-by-step explanation:

Let's sketch graphs of functions f(x) and g(x) on one coordinate system (attachment).

Let's calculate the common points:

$x^2=\sqrt{x}\qquad\text{square of both sides}\\\\(x^2)^2=\left(\sqrt{x}\right)^2\\\\x^4=x\qquad\text{subtract}\ x\ \text{from both sides}\\\\x^4-x=0\qquad\text{distribute}\\\\x(x^3-1)=0\iff x=0\ \vee\ x^3-1=0\\\\x^3-1=0\qquad\text{add 1 to both sides}\\\\x^3=1\to x=\sqrt[3]1\to x=1$

The area to be calculated is the area in the interval [0, 1] bounded by the graph g(x) and the axis x minus the area bounded by the graph f(x) and the axis x.

We have integrals:

$\int\limits_{0}^1(\sqrt{x})dx-\int\limits_{0}^1(x^2)dx=(*)\\\\\int(\sqrt{x})dx=\int\left(x^\frac{1}{2}\right)dx=\dfrac{2}{3}x^\frac{3}{2}=\dfrac{2x\sqrt{x}}{3}\\\\\int(x^2)dx=\dfrac{1}{3}x^3\\\\(*)=\left(\dfrac{2x\sqrt{x}}{2}\right]^1_0-\left(\dfrac{1}{3}x^3\right]^1_0=\dfrac{2(1)\sqrt{1}}{2}-\dfrac{2(0)\sqrt{0}}{2}-\left(\dfrac{1}{3}(1)^3-\dfrac{1}{3}(0)^3\right)\\\\=\dfrac{2(1)(1)}{2}-\dfrac{2(0)(0)}{2}-\dfrac{1}{3}(1)}+\dfrac{1}{3}(0)=2-0-\dfrac{1}{3}+0=1\dfrac{1}{3}$

6 0

3 years ago