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zaharov [31]
3 years ago
11

How many triangles can be constructed using the following three side lengths: 4.8 cm, 2.1 cm, and 2.6 cm?

Mathematics
1 answer:
Colt1911 [192]3 years ago
5 0

Answer: No triangle

Step-by-step explanation:

The given dimensions of triangle : 4.8 cm, 2.1 cm, and 2.6 cm

  • The famous triangle inequality says that in any triangle the sum of the lengths of any two sides must be greater than or equal to the length of the remaining side.

But 2.1+2.6=4.7

It means the sum of two sides (2.1 and 2.6) is less than the third side (4.8).

Therefore by triangle inequality, no triangle can be possible by the given measurement.

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What does X equal in the equation 8(4-x) =7x +2
drek231 [11]

Answer:

x = 2

Step-by-step explanation:

Given

8(4 - x) = 7x + 2 ← distribute parenthesis on left side

32 - 8x = 7x + 2 ( subtract 7x from both sides )

32 - 15x = 2 ( subtract 32 from both sides )

- 15x = - 30 ( divide both sides by - 15 )

x = 2

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3 years ago
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Jada was using square stickers that were 3/4in long to decorate the spine of a photo album. The spine is 10 and 1/2in long. If s
nevsk [136]

Answer:

13

Step-by-step explanation:

Given:

  1. Square stickers that were 3/4 in long
  2. The spine is 10 and 1/2 in long

If she laid the stickers side by side without gaps or overlaps so the number of sticker to cover the length of the spine:

\frac{The length of the spine}{the length of the sticker}

10 : \frac{3}{4} ≈13  

She needs 13 stickers cover the length of the spine

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5% tax on a $42.00 purchase under $2.00?
DaniilM [7]
Divide the 5 with 42.00 and see the answer u got
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Convert 5 miles to inches
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If (x^2 - 1) is a factor of ax^4 + bx^3 + cx^2 + dx + e, show that a + c + e = b + d = 0
alisha [4.7K]
x^2-1=x^2-1^2=(x-1)(x+1)

If x²-1 is a factor of the polynomial, both x-1 and x+1 are factors of it.

According to the remainder theorem, if a binomial x-a is a factor of a polynomial p(x), then p(a)=0.

If x-1 and x+1 are factors of the polynomial p(x)=ax⁴+bx³+cx²+dx+e, then p(1)=0 and p(-1)=0.

p(1)=a \times 1^4+b \times 1^3 + c \times 1^2+ d \times 1+e \\
p(-1)=a \times (-1)^4 + b \times (-1)^3 + c \times (-1)^2 + d \times (-1)+e \\ \\
p(1)=a+b+c+d+e \\
p(-1)=a-b+c-d+e \\ \\
p(1)=0 \\
p(-1)=0 \\ \\ \hbox{add both equations:} \\
a+b+c+d+e=0 \\
\underline{a-b+c-d+e=0} \\
2a+2c+2e=0 \\
2(a+c+e)=0 \\
a+c+e=0 \\ \\
\hbox{substitute 0 for a+c+e in the first equation:} \\
a+b+c+d+e=0 \\
(a+c+e)+b+d=0 \\
0+b+d=0 \\
b+d=0 \\ \\
\boxed{a+c+e=b+d=0} \\
\hbox{proved } \checkmark
8 0
3 years ago
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