All categories

2 years ago

I WILL GIVE 20 POINTS TO THOSE WHO ANSWER THIS QUESTION RIGHT NOOOO SCAMS AND EXPLAIN WHY THAT IS THE ANSWER

Mathematics

1 answer:

Leya [2.2K]2 years ago

8 0

Step-by-step explanation:

the volume of a regular prism is simply

base area × height

in our case we have a triangular prism, which means the base area (as well as the top area) is a triangle.

the base is defined as a 45° - 45° - 90° triangle.

so, since we have a 90° angle, this is a right-angled triangle.

the other 2 angles are equal to each other, so, it is an isoceles triangle (both legs are equally long).

since one leg is 8 in long, this means the other leg is 8 in too.

the area of a triangle is

baseline × height / 2

in case of a right-angled triangle (the legs enclose the 90° angle) one leg can be considered as the baseline, and the other leg as the height.

and the area is

leg1 × leg2 / 2

and in case of an isoceles right-angled triangle this is

leg² / 2

in our case

8² / 2 = 64/2 = 32 in²

now, the volume of the prism here is then

32 × 3.5 = 112 in³

You might be interested in

URGENT!!!! WILL GIVE BRAINLIEST TO FIRST RIGHT ANSWER!!!!!!

Zepler [3.9K]

geometric sequence

$\tt U_n=ar^{n-1}$

a = 6 = first term

r = ratio = $\tt \dfrac{9}{6}$

the 6th term (n=6)

$\tt U_6=6\times (\dfrac{9}{6})^{6-1}\\\\U_6=6\times \dfrac{9}{6}^5\\\\U_6=\dfrac{9^5}{6^4}=45.5625\rightarrow rounded~to nearest~thousandth=45.563$

4 0

2 years ago

In the triangle pictured, let A, B, C be the angles at the three vertices, and let a,b,c be the sides opposite those angles. Acc

Troyanec [42]

Answer:

Step-by-step explanation:

(a)

Consider the following:

$A=\frac{\pi}{4}=45°\\\\B=\frac{\pi}{3}=60°$

Use sine rule,

$\frac{b}{a}=\frac{\sinB}{\sin A}
\\\\=\frac{\sin{\frac{\pi}{3}}
}{\sin{\frac{\pi}{4}}}\\\\=\frac{[\frac{\sqrt{3}}{2}]}{\frac{1}{\sqrt{2}}}\\\\=\frac{\sqrt{2}}{2}\times \frac{\sqrt{2}}{1}=\sqrt{\frac{3}{2}}$

Again consider,

$\frac{b}{a}=\frac{\sin{B}}{\sin{A}}
\\\\\sin{B}=\frac{b}{a}\times \sin{A}\\\\\sin{B}=\sqrt{\frac{3}{2}}\sin {A}\\\\B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]$

Thus, the angle B is function of A is, $B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]$

Now find $\frac{dB}{dA}$

Differentiate implicitly the function $\sin{B}=\sqrt{\frac{3}{2}}\sin{A}$ with respect to A to get,

$\cos {B}.\frac{dB}{dA}=\sqrt{\frac{3}{2}}\cos A\\\\\frac{dB}{dA}=\sqrt{\frac{3}{2}}.\frac{\cos A}{\cos B}$

When $A=\frac{\pi}{4},B=\frac{\pi}{3}$ , the value of $\frac{dB}{dA}$ is,

$\frac{dB}{dA}=\sqrt{\frac{3}{2}}.\frac{\cos {\frac{\pi}{4}}}{\cos {\frac{\pi}{3}}}\\\\=\sqrt{\frac{3}{2}}.\frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}}\\\\=\sqrt{3}$

In general, the linear approximation at x= a is,

$f(x)=f'(x).(x-a)+f(a)$

Here the function $f(A)=B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]$

At $A=\frac{\pi}{4}$

$f(\frac{\pi}{4})=B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{\frac{\pi}{4}}]\\\\=\sin^{-1}[\sqrt{\frac{3}{2}}.\frac{1}{\sqrt{2}}]\\\\\=\sin^{-1}(\frac{\sqrt{2}}{2})\\\\=\frac{\pi}{3}$