All categories

3 years ago

An angle measures 80° less than the measure of its supplementary angle. What is the measure of each angle?

Mathematics

1 answer:

Lostsunrise [7]3 years ago

5 0

Answer:

50° and 130°

Step-by-step explanation:

let one angle be x then the other angle is x - 80

Supplementary angles sum to 180°, thus

x + x - 80 = 180, that is

2x - 80 = 180 ( add 80 to both sides )

2x = 260 ( divide both sides by 2 )

x = 130

The angles are 130° and 130 - 80 = 50°

You might be interested in

EZ 15 points!

nikdorinn [45]

Answer:

In a collision between two objects, both objects experience forces that are equal in magnitude and opposite in direction. Thus, if the colliding objects have unequal mass, they will have unequal accelerations as a result of the contact force that results during the collision.

Step-by-step explanation:

dont know the question 2 answer

hope the first one helps tho

3 0

3 years ago

The ratio of sugar to flour in the recipe is 1 to

Sveta_85 [38]

Answer:

10 flour

Step-by-step explanation:

1:2

5:x all you have to do is multiply by 2, sense the ratio is 1:2, flour will always be 2 times greater than sugar. so the answer is 5 x 2 =10

5 0

3 years ago

8,100 divided by 9= 90. Yes or No?

madam [21]

No
Because, 8,100÷9=900

5 0

4 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

Question 1

lesya [120]

It would be 12h :))((:

7 0

3 years ago