Ask question

Ask question

All categories

2 years ago

12

14. There are 6000 people at an ice hockey match. The announcer says this is exactly 40% more people that the previous match. Ex

plain why the announcer is incorrect

1 answer:

nata0808 [166]2 years ago

7 0

Answer:

because he said that in place of than

You might be interested in

Find both X's Please show your work!

Hoochie [10]

Right angle=90°.
(X-20) + X = 90.
Then solve for X.
X=55

4 0

2 years ago

I need help with this plzzzzzzzzzzzz

Sunny_sXe [5.5K]

Answer:

6ft by 15ft

Step-by-step explanation:

just divide each number by two and add it to the origanal number.

7 0

3 years ago

Khan Academy Order Rational Numbers<br><br> Please help, I will mark brainliest.

11111nata11111 [884]

Answer:

C= -1/5

B= -0.4

C= -3/7

Step-by-step explanation:

6 0

3 years ago

The point (1/3,1/4) lies on the terminal said of an angle. Find the exact value of the six trig functions and explain which func

katrin2010 [14]

Answer:

sine and cosec are inverse of each other.

cosine and sec are inverse of each other.

tan and cot are inverse of each other.

Step-by-step explanation:

Given point on terminal side of an angle ( $\frac{1}{3},\frac{1}4$ ).

Kindly refer to the attached image for the diagram of the given point.

Let it be point A( $\frac{1}{3},\frac{1}4$ )

Let O be the origin i.e. (0,0)

Point B will be ( $\frac{1}{3},0$ )

Now, let us consider the right angled triangle $\triangle OBA$ :

Sides:

$Base, OB = \frac{1}{3}\\Perpendicular, AB = \frac{1}{4}$

Using Pythagorean theorem:

$\text{Hypotenuse}^{2} = \text{Base}^{2} + \text{Perpendicular}^{2}\\\Rightarrow OA^{2} = OB^{2} + AB^{2}\\\Rightarrow OA^{2} = \frac{1}{3}^{2} + \frac{1}{4}^{2}\\\Rightarrow OA = \sqrt{\frac{1}{3}^{2} + \frac{1}{4}^{2}}\\\Rightarrow OA = \sqrt{\frac{4^2+3^2}{3^{2}.4^2 }}\\\Rightarrow OA = \frac{5}{12}$

$sin \angle AOB = \dfrac{Perpendicular}{Hypotenuse}$

$\Rightarrow sin \angle AOB = \dfrac{\frac{1}{4}}{\frac{5}{12}}\\\Rightarrow sin \angle AOB = \dfrac{3}{5}$

$cos\angle AOB = \dfrac{Base}{Hypotenuse}$

$\Rightarrow cos \angle AOB = \dfrac{\frac{1}{3}}{\frac{5}{12}}\\\Rightarrow cos\angle AOB = \dfrac{4}{5}$

$tan\angle AOB = \dfrac{Perpendicular}{Base}$

$\Rightarrow tan\angle AOB = \dfrac{3}{4}$

$cosec \angle AOB = \dfrac{Hypotenuse}{Perpendicular}$

$\Rightarrow cosec\angle AOB = \dfrac{5}{3}$

$sec\angle AOB = \dfrac{Hypotenuse}{Base}$

$\Rightarrow sec\angle AOB = \dfrac{5}{4}$

$cot\angle AOB = \dfrac{Base}{Perpendicular}$

$\Rightarrow cot\angle AOB = \dfrac{4}{3}$

3 0

3 years ago

Legs and hypotenuse

zaharov [31]

Answer:

Step-by-step explanation:

For a right triangle, the side that is opposite of the right angle is called the hypotenuse. This side will always be the longest side of the right triangle. The other two (shorter) sides are called legs.

3 0

3 years ago