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3 years ago

Find the surface area of a triangular prism

Mathematics

2 answers:

MrRissso [65]3 years ago

8 0

That’s the answer. Dhjahsgs

Elan Coil [88]3 years ago

3 0

$\huge\mathrm{Answer࿐}$

Surface area of Triangular prism :

$\mathrm{(2 \times \dfrac{1}{2} \times 8 \times 6) + (3 \times 10 \times 20)}$

$\mathrm{48 +600 }$

$648$

Surface area = 648 cm²

_____________________________

$\mathrm{ \#TeeNForeveR}$

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2 years ago

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Answer:

Dinda received a 32% loss percentage.

Step-by-step explanation:

<u><em>The question in English is</em></u>

Dinda bought a phone for Rp.2,500,000.00. After 5 months, she sells it with

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we know that

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Using a proportion

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6 0

3 years ago

How do u work this problem out

charle [14.2K]

Okay what problem is it that you need help with??

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4 years ago

Robin had 7/8 pound of butter when she started baking a cake. She used 1/4 pound of butter while baking. How much butter was lef

8_murik_8 [283]

Answer:

$\dfrac{5}{8}$ butter was left after she was done baking.

Step-by-step explanation:

Robin had 7/8 pound of butter when she started baking a cake. She used 1/4 pound of butter while baking.

We need to find how much butter was left after she was done baking.

It can be calculated as follows,

Left =

$=\dfrac{7}{8}-\dfrac{1}{4}\\\\=\dfrac{7}{8}-\dfrac{2}{8}\\\\=\dfrac{7-2}{8}\\\\=\dfrac{5}{8}$

Hence, $\dfrac{5}{8}$ butter was left after she was done baking

3 0

3 years ago

The coordinates of three of the vertices of parallelogram ABCD are A(1, 0), B(2, 3), C(3, 2). What are coordinates of the fourth

nalin [4]

Answer:

A) The coordinates of the fourth vertex are:

1) x-coordinate:

$x=2$

2) y-coordinate:

$y=-1$

B) The point of intersection of the diagonals is: $(2,1)$

Step-by-step explanation:

We need to remember that the diagonals of a parallelogram intersect each other at a half-way point and the midpoint of each diagonal is the same.

The midpoint formula is:

$M=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

Since:

$M_{AC}=M_{BD}$

We can find the coordinates of the fourth vertex $D(x,y)$ through these procedure:

1) x-coordinate:

$\frac{1+3}{2}=\frac{2+x}{2}\\\\2(2)=x\\\\4-2=x\\\\x=2$

2) y-coordinate:

$\frac{0+2}{2}=\frac{3+y}{2}\\\\1(2)-3=y\\\\y=-1$

Therefore, fourth vertex is $D(2,-1)$

Since the point of intersection of the diagonals is the midpoint of a diagonal (Remember that $M_{AC}=M_{BD}$ ), this is:

$M_{AC}=M_{BD}=(\frac{1+3}{2},\frac{0+2}{2})\\\\M_{AC}=M_{BD}=(2,1)$

Therefore, the point of intersection of the diagonals is $(2,1)$

7 0

4 years ago