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3 years ago

7

Help me plz :3 :) :x :P

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