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kolbaska11 [484]
3 years ago
11

What is the volume of the figure?

Mathematics
1 answer:
Gelneren [198K]3 years ago
5 0

144 cubic inches

Step-by-step explanation:

V=lwh

6x4=24

24x6=144

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What is the total surface area
Stels [109]

We have two right triangles and three different rectangles.

The formula of an area of a right triangle:

A_T=\dfrac{1}{2}l_1l_2

l₁, l₂ - legs

We have l₁ = 20cm and l₂ = 21cm. Substitute:

A_T=\dfrac{1}{2}(20)(21)=(10)(21)=210\ cm^2

The formula of an area of a rectangle:

A_R=lw

l - length

w - width

We have:

rectangle #1: l = 22cm, w = 29cm

A_{R1}=(22)(29)=638\ cm^2

rectangle #2: l = 22cm, w = 21cm

A_{R2}=(22)(21)=462\ cm^2

rectangle #3: l = 22cm, w = 20cm

A_{R3}=(22)(20)=440\ cm^2

The total Surface Area of the triangular prism:

S.A.=2A_T+A_{R1}+A_{R2}+A_{R3}\\\\S.A.=2\cdot210+638+462+440=1960\ cm^2

3 0
3 years ago
Use the method of completing the square to solve x2 - 6x + 1 = 0. Simplify your solutions.
exis [7]

Answer:

x = 3 ± 2\sqrt{2}

Step-by-step explanation:

given

x² - 6x + 1 = 0 ( subtract 1 from both sides )

x² - 6x = - 1

To complete the square

add (half the coefficient of the x- term)² to both sides

x² - 6x + (- 3)² = - 1 + (- 3)²

x² - 6x + 9 = - 1 + 9

(x - 3)² = 8 ( take the square root of both sides )

x - 3 = ± \sqrt{8} = ± 2\sqrt{2} ( add 3 to both sides )

x = 3 ± 2\sqrt{2}

8 0
3 years ago
Can someone help me with this ?!! Plz
Natasha2012 [34]

Answer:

see image

Step-by-step explanation:

8 0
3 years ago
(w-x)^2 + 26x; if w = 6 and x = -1
kati45 [8]

Answer:

\boxed{\boxed{\sf 23}}

Step-by-step explanation:

\boxed{\sf Hi\: there!}

\sf (w-x)^2 + 26x

\sf w=6

\sf x=-1

_____________________

→ \sf \left(6-\left(-1\right)\right)^2+26\left(-1\right)

→ \sf 23

»»————- ➴ ————-««

7 0
2 years ago
Read 2 more answers
T F IfA and B are similar matrices, then AT=BT
muminat

Answer:

Step-by-step explanation:

We know that for two similar matrices A and B exists an invertible matrix P for which

[tex]B = P^{-1} AP[/tex]

∴ B^{T} = (P^{-1})^{T} A^{T} P^{T} \\

Also P^{-1}P = I\\

and (P^{-1})^{T} = (P^{T})^{-1}

∴(P^{-1})^{T}P^{T} = I

so, B^{T} = (P^{-1})^{T} A^{T} P^{T} = (P^{T})^{-1}A^{T} P^{T}\\B^{T} = A^{T} I\\B^{T} = A^{T}

7 0
3 years ago
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