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lesya692 [45]
2 years ago
7

Find the collame of boid of dimension 16cm X 10m x 6cm.​

Mathematics
1 answer:
swat322 years ago
7 0

Question: Find the volume of cuboid of dimension 16cm X 10cm x 6cm.

Answer: Given Length=16cm, Breadth=10cm, Height=6cm

Using Volume formula for cuboid:-

volume = l \times b \times h

→16 \times 10 \times 6

→960 \: cm^{3}

So, volume of cuboid is <u>9</u><u>6</u><u>0</u><u> </u><u>cm³</u>

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I think it’s 3 not sure tho because i forgot most things about it
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If ABC forms a linear pair with CBD, use the Linear Pair Postulate to determine the value of x
arsen [322]
There you go any questions?

7 0
2 years ago
Someone please help me
kap26 [50]

Answer:

The answer is x = 4

Step-by-step explanation:

1. First you need to distribute the 3/4 * (x + 8). This looks like (3/4) * (x) + (3/4) * (8) = 9

2. Next you simplify the distributed equation, 3/4x + 6 = 9

3. Now subtract 6 from both sides, 3/4x = 3

4. Multiply both sides by 4/3, 4/x * 3/4x = 3 * 4/3

5. Simplify, x = 4

8 0
3 years ago
X - 2y = 3<br> 5x + 3y = 2<br> The lines whose equations are shown intersect at which point?
Lady_Fox [76]

The point of intersection of the two lines is at (1,-1)

<h3>System of equation</h3>

The given system of expression is shown below

x - 2y = 3

5x + 3y = 2

The solution to the system of equation is the point of intersection

From equation 1

x = 3 + 2y

Substitute into 2

5(3+2y) + 3y = 2

15 +10y + 3y = 2

13y = -13

y = -1

Substitute y = -1 into 3

x = 3 + 2y

x = 3+(-2)

x = 1

Hence the point of intersection of the two lines is at (1,-1)

Learn more on system of equation here: brainly.com/question/25976025

#SPJ1

8 0
2 years ago
PLEASE HELP... VERY URGENT... WILL GIVE BRAINLIEST!!!
lakkis [162]

Answer:

Step-by-step explanation:

a) Pythagorean theorem,

altitude² + base² = hypotenuse²

h²  + (w² - 1)² = (w² + 1)²

                    h² = (w² + 1)² - (w² - 1)²    

{Compare with (a + b)² - (a -b)² = 4ab where a =w² & b = 1}

                   h² = 4*w²*1

                   h² = 4w²

                   h=\sqrt{4w^{2}} = \sqrt{2*2*w*w} \\\\\\h = 2w

b) Area of triangle = \frac{1}{2}b*h

                               =\dfrac{1}{2}*(w^{2}-1) *2w\\\\= (w^{2}-1)*w = w^{2}*w - 1*w\\\\=w^{3}-w

c) w =2

Area of triangle = 3w = 3*2 = 6

7 0
2 years ago
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