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3 years ago

What is the surface area of the triangular prism shown

Mathematics

1 answer:

Zigmanuir [339]3 years ago

3 0

The surface area of the shape shown is 1008 square centimeters. I hope this helps!

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-14.9=x/8 -2.1 Please help me answer

Romashka [77]

-14.9 = $\frac{x}{8}$ - 2.1 (Add 2.1 to both sides)
-12.8 = $\frac{x}{8}$ (Multiply both sides by 8)
-102.4 = x

~~~Hope this helps!~~~

4 0

3 years ago

Find the first five terms of the sequence defined below, where n represents the position of a

SOVA2 [1]

Answer:

Your answer is going to be 4, 7, 10, 13, 16

Step-by-step explanation:

n=1, 3*1+1=4

n=2, 3*2+1=7

n=3, 3*3+1=10

n=4, 3*4+1=13

n=5, 3*5+1=16

6 0

3 years ago

Graph y=-3x +7 and the line that is parallel to it that passes through (4,1).

Brums [2.3K]

Y = -3x + 13

Steps and the graph are on paper

6 0

3 years ago

You are driving to a conference in Cleveland and have already traveled 100 miles. You still have 50 more miles to go. When you a

Alborosie

Answer:

150 miles

Step-by-step explanation:

If you go 50 more miles after you have gone 100, you have traveled a total of ...

100 + 50 = 150 . . . miles

6 0

3 years ago

Read 2 more answers

Prove that: (b²-c²/a)CosA+(c²-a²/b)CosB+(a²-b²/c)CosC = 0

IRISSAK [1]

Prove that:

$\:\:\sf\:\:\left(\dfrac{b^2-c^2}{a}\right)\cos A+\left(\dfrac{c^2-a^2}{b}\right)\cos B +\left(\dfrac{a^2-b^2}{c}\right)\cos C=0$

Proof: 

We know that, by Law of Cosines,

$\sf \cos A=\dfrac{b^2+c^2-a^2}{2bc}$
$\sf \cos B=\dfrac{c^2+a^2-b^2}{2ca}$
$\sf \cos C=\dfrac{a^2+b^2-c^2}{2ab}$

Taking LHS

$\left(\dfrac{b^2-c^2}{a}\right)\cos A+\left(\dfrac{c^2-a^2}{b}\right)\cos B +\left(\dfrac{a^2-b^2}{c}\right)\cos C$

Substituting the value of cos A, cos B and cos C,

$\longmapsto\left(\dfrac{b^2-c^2}{a}\right)\left(\dfrac{b^2+c^2-a^2}{2bc}\right)+\left(\dfrac{c^2-a^2}{b}\right)\left(\dfrac{c^2+a^2-b^2}{2ca}\right)+\left(\dfrac{a^2-b^2}{c}\right)\left(\dfrac{a^2+b^2-c^2}{2ab}\right)$

$\longmapsto\left(\dfrac{(b^2-c^2)(b^2+c^2-a^2)}{2abc}\right)+\left(\dfrac{(c^2-a^2)(c^2+a^2-b^2)}{2abc}\right)+\left(\dfrac{(a^2-b^2)(a^2+b^2-c^2)}{2abc}\right)$

$\longmapsto\left(\dfrac{(b^2-c^2)(b^2+c^2)-(b^2-c^2)(a^2)}{2abc}\right)+\left(\dfrac{(c^2-a^2)(c^2+a^2)-(c^2-a^2)(b^2)}{2abc}\right)+\left(\dfrac{(a^2-b^2)(a^2+b^2)-(a^2-b^2)(c^2)}{2abc}\right)$

$\longmapsto\left(\dfrac{(b^4-c^4)-(a^2b^2-a^2c^2)}{2abc}\right)+\left(\dfrac{(c^4-a^4)-(b^2c^2-a^2b^2)}{2abc}\right)+\left(\dfrac{(a^4-b^4)-(a^2c^2-b^2c^2)}{2abc}\right)$