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2 years ago

7

If 20 articles cost ? 90, then the cost of 9 articles is ? ? 45 ? 4.50 ? 40.50 ? 42

1 answer:

lapo4ka [179]2 years ago

5 0

Answer:

40.50 is correct answer according to me

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What is the area of this trapezoid?<br><br><br><br> Enter your answer in the box.

ANEK [815]

Step-by-step explanation:

a=7+3+7=17

b=3

h=8

area=(1/2)(17+3)(8)=80

5 0

2 years ago

Read 2 more answers

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

2 years ago

The following table shows values from a linear function. What is the

Naddik [55]

Answer:

9

Step-by-step explanation:

The equation graphed from these points is y=3x+9, therefore the y-intercept is 9.

4 0

2 years ago

Prove that:- (1-sinA)(1+sinA)÷(1+cosA)(1-cosA)=cot^2A

astraxan [27]

Answer:

see explanation

Step-by-step explanation:

Using the trigonometric identity

sin²x + cos²x = 1 , then

sin²x = 1 - cos²x and cos²x = 1 - sin²x

Consider the left side

$\frac{(1-sinA)(1 + sinA)}{(1+cosA)(1-cosA)}$ ← expand numerator/denominator using FOIL

= $\frac{1-sin^2A}{1-cos^2A}$

= $\frac{1-(1-cos^2A)}{1-(1-sin^2A)}$

= $\frac{1-1+cos^2A}{1-1+sin^2A}$

= $\frac{cos^2A}{sin^2A}$

= cot²A = right side , thus proven

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2 years ago

Distribute and simplify these radicals. 23-(V+ 3) O 2V5 + 6 02/15 O 30 O 6/2+6

zhannawk [14.2K]

Answer:

A)

Step-by-step explanation:

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3 years ago