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lorasvet [3.4K]

2 years ago

11

Determine the surface area of a square based pyramid that has a slant height of 6 in and a base length of 8 in? Round your answe

r to the nearest whole.

2 answers:

likoan [24]2 years ago

7 0

Answer: 48

6x8 calculate the product or quotient

48

damaskus [11]2 years ago

3 0

Answer:

48

Step-by-step explanation:

6x8 for slant height and base length gives you 48

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Compare the light gathering power of an 8" primary mirror with a 6" primary mirror. The 8" mirror has how much light gathering p

tankabanditka [31]

Answer:

1.778 times more or 16/9 times more

Step-by-step explanation:

Given:

- Mirror 1: D_1 = 8''

- Mirror 2: D_2 = 6"

Find:

Compare the light gathering power of an 8" primary mirror with a 6" primary mirror. The 8" mirror has how much light gathering power?

Solution:

- The light gathering power of a mirror (LGP) is proportional to the Area of the objects:

LGP ∝ A

- Whereas, Area is proportional to the squared of the diameter i.e an area of a circle:

A ∝ D^2

- Hence, LGP ∝ D^2

- Now compare the two diameters given:

LGP_1 ∝ (D_1)^2

LGP ∝ (D_2)^2

- Take a ratio of both:

LGP_1/LGP_2 ∝ (D_1)^2 / (D_2)^2

- Plug in the values:

LGP_1/LGP_2 ∝ (8)^2 / (6)^2

- Compute: LGP_1/LGP_2 ∝ 16/9 ≅ 1.778 times more

6 0

3 years ago

What is the area of a parallelogram that has a base of 12 and 3 fourths and a height of 2 and 1 half

storchak [24]

Area of Parallelogram= base * height
A= (12 3/4)(2 1/2)
A= (51/4)(5/2)
A= 255/8
A= 31 7/8

OR
A= (12.75)(2.5)
A= 31.875

Hope this helps! :)

6 0

4 years ago

By selling a watch for Rs. 150, there would be a loss and there would be a proft if it is sold for Rs. 200. If the loss and the

Nuetrik [128]

Rs. 25 loss

Rs. 25 profit

AND Rs. 150 cost.

6 0

2 years ago

Which of the following is a solution to 2cos2x − cos x − 1 = 0?

Pavel [41]

Answer:

Option A is correct.

Solution for the given equation is, $x = 0^{\circ}$

Step-by-step explanation:

Given that : $2\cos^2x -\cos x -1 =0$

Let $\cos x =y$

then our equation become;

$2y^2-y-1= 0$ .....[1]

A quadratic equation is of the form:

$ax^2+bx+c =0$ .....[2] where a, b and c are coefficient and the solution is given by;

$x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Comparing equation [1] and [2] we get;

a = 2 b = -1 and c =-1

then;

$y = \frac{-(-1)\pm \sqrt{(-1)^2-4(2)(-1)}}{2(2)}$

Simplify:

$y = \frac{ 1 \pm \sqrt{1+8}}{4}$

or

$y = \frac{ 1 \pm \sqrt{9}}{4}$

$y = \frac{ 1 \pm 3}{4}$

or

$y = \frac{1+3}{4}$ and $y = \frac{1 -3}{4}$

Simplify:

y = 1 and $y = -\frac{1}{2}$

Substitute y = cos x we have;

$\cos x = 1$

⇒ $x = 0^{\circ}$

and

$\cos x = -\frac{1}{2}$

⇒ $x = 120^{\circ} \text{and} x = 240^{\circ}$

The solution set: $\{0^{\circ}, 120^{\circ} , 240^{\circ}\}$

Therefore, the solution for the given equation $2\cos^2x -\cos x -1 =0$ is, $0^{\circ}$

8 0

4 years ago

Read 2 more answers

13 NEED HELP ASAP WILL GIVE BRAINLY

attashe74 [19]

Answer:

A. Similar; congruent angles

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers