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kherson [118]
3 years ago
7

Help me plz :3 :) :x :P

Mathematics
1 answer:
Sonbull [250]3 years ago
7 0
Yes, she is correct
1/2 + 1/2 = 1 or 0.5 + 0.5 = 1 so half of a wall plus half of a wall equals one hole wall.

That's as simple as I can put it mathematical reasoning. I hope this helps
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BRAINLIEST!!! HELP!!<br> 3. Rotate the triangle counterclockwise about point D.
Ksivusya [100]
The answer is the first one
8 0
3 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
3 years ago
I need help on #7<br> Please show your work
givi [52]

The area of the given square pyramid is:

total area = 1,100 inches squared.

<h3 /><h3>How to get the area of the pyramid?</h3>

On the second image, we can see that the pyramid is conformed of a square base and 3 triangles.

To get the surface area of the pyramid, we can just get the area of each of these simpler parts.

The base is a square of 22 in by 22 in, then the area of the base is:

B = (22in)*(22 in)  = 484 in^2

For each triangle, the area will be:

A = (base side)*(height)/2

A = (22in)*(14in)/2 = 154 in^2

And we have 4 of these triangles, then the total area of the pyramid will be:

total area = B + 4*A = 484in^2 + 4*(154 in^2) = 1,100 in^2

If you want to learn more about square pyramids:

brainly.com/question/22744289

#SPJ1

3 0
1 year ago
4<br> Om obtuse, scalene<br> Obright, scalene<br> Oc, right, isosceles<br> Od, right, equilateral
harina [27]

Answer:

B

Step-by-step explanation:

B. right, scalene

3 0
2 years ago
Mark the points A(5, 5) and B(7.2). Draw a circle that has point A as the center and passes through point B. Measure and label t
olasank [31]

Answer:

i think 4 units im not sure though though

Step-by-step explanation:

4 0
3 years ago
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