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Fantom [35]
2 years ago
11

Find the volume of the irregular figure'

Mathematics
1 answer:
GrogVix [38]2 years ago
3 0

Answer:

136 cm³

Step-by-step explanation:

<u>Volume of the figure:-</u>

  • (6×4×5)+(4×2×2) cm³
  • 120+16 cm³
  • 136 cm³

hope it's helpful...

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I need help with this proof
Maksim231197 [3]

Answer:

See explanation

Step-by-step explanation:

Consider two triangles ABF and EDF. These triangles are two right triangles, because angles B and D are right anges.

In these triangles,

  • ∠ABF ≅ ∠EDF - as two right angles;
  • BF ≅ FE - given;
  • ∠AFB ≅ ∠EFD - as vertical angles when two lines AD and BE itersect.

By ASA postulate (or HA postulate) these triangles are congruent, so

ΔABF ≅ ΔEDF

Congruent triangles have congruent corresponding parts, so

FA ≅ EF

6 0
3 years ago
Explain what the number 0 on the gauge represents and explain what the numbers above 0 represent
DaniilM [7]
0 is 500mg and the numbers above are larger.
3 0
3 years ago
If EFGH is a rectangle, what is FH?
lys-0071 [83]
I believe FH would be a line segment.
Hope this helps!
3 0
3 years ago
2x3 + x2 + 7x - 6)-(- 2x + 10x + 10x+9)​
AVprozaik [17]

Answer:

2x³ + x² - 11x - 15

Step-by-step explanation:

Step 1: Write out expression

2x³ + x² + 7x - 6 - (-2x + 10x + 10x + 9)

Step 2: Distribute negative

2x³ + x² + 7x - 6 + 2x - 10x - 10x - 9

Step 3: Combine like terms (x)

2x³ + x² + 9x - 10x - 10x - 9 - 6

2x³ + x² - x - 10x - 9 - 6

2x³ + x² - 11x - 9 - 6

Step 4: Combine like terms (constants)

2x³ + x² - 11x - 15

5 0
3 years ago
A triangle is graphed in the coordinate plane. The vertices of the triangle have coordinates (–3, 1), (1, 1), and (1, –2). What
vladimir2022 [97]

Answer:

The perimeter of the triangle is 12\ units

Step-by-step explanation:

Let

A(-3,1),B(1,1),C(1,-2)

we know that

The perimeter of triangle is equal to

P=AB+BC+AC

the formula to calculate the distance between two points is equal to

d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}

step 1

Find the distance AB

A(-3,1),B(1,1)

substitute in the formula

AB=\sqrt{(1-1)^{2}+(1+3)^{2}}

AB=\sqrt{(0)^{2}+(4)^{2}}

AB=4\ units

step 2

Find the distance BC

B(1,1),C(1,-2)

substitute in the formula

BC=\sqrt{(-2-1)^{2}+(1-1)^{2}}

BC=\sqrt{(-3)^{2}+(0)^{2}}

BC=3\ units

step 3

Find the distance AC

A(-3,1),C(1,-2)

substitute in the formula

AC=\sqrt{(-2-1)^{2}+(1+3)^{2}}

AC=\sqrt{(-3)^{2}+(4)^{2}}

AC=5\ units

step 4

Find the perimeter

P=AB+BC+AC

substitute the values

P=4+3+5=12\ units

6 0
3 years ago
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