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myrzilka [38]
3 years ago
5

Jensen wants to hang a mirror in his boat and put a frame around it. The mirror and frame must have an area of 19.75 square feet

. The mirror is 4 feet wide and 5 feet long. Which quadratic equation can be used to determine the thickness of the frame, x? (2 points)
2x2 + 6x − 19.25 = 0

3x2 + 5x − 19.25 = 0

4x2 + 18x + 0.25 = 0

4x2 + 18x + 20 = 0
Mathematics
1 answer:
Ber [7]3 years ago
6 0
The third choice, 4x²+18x+0.25=0, is correct.

The frame will have the same thickness on both sides of the mirror.  Therefore we add 5+x+x to get the total length, and we have 4+x+x to get the total width. 

The area is found by multiplying the total width and total length:

A=(5+x+x)(4+x+x)
A=(5+2x)(4+2x)

Multiplying the binomials, we have:
A=5*4+5*2x+2x*4+2x*2x
A=20+10x+8x+4x²
A=20+18x+4x²

We know that the area is 19.75:
19.75=20+18x+4x²

We need quadratics set equal to 0, so subtract 19.75 from each side:
19.75-19.75 = 20+18x+4x²-19.75
0=0.25+18x+4x²


In standard form, it will be:
4x²+18x+0.25=0
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Answer:

The value of T₂₀ - T₁₅ is <u>-20</u>.

Step-by-step explanation:

<u>Given</u> :

  • >> If for an A.P, d = -4

<u>To</u><u> </u><u>Find</u> :

  • >> T₂₀ - T₁₅

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\star{\small{\underline{\boxed{\sf{\red{ T_n = a  + (n - 1)d}}}}}}

  • >> Tₙ = nᵗʰ term
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<u>Solution</u> :

Firstly finding the A.P of T₂₀ by substituting the values in the formula :

{\dashrightarrow{\pmb{\sf{ T_n = a  + (n - 1)d}}}}

{\dashrightarrow{\sf{ T_{20} = a  + (20 - 1) d}}}

{\dashrightarrow{\sf{ T_{20} = a  + (19)d}}}

{\dashrightarrow{\sf{ T_{20} = a  + 19  \times d}}}

{\dashrightarrow{\sf{ T_{20} = a  + 19d}}}

{\star \: {\underline{\boxed{\sf{\pink{ T_{20} = a  + 19d}}}}}}

Hence, the value of T₂₀ is a + 19d.

\rule{190}1

Secondly, finding the A.P of T₁₅ by substituting the values in the formula :

{\dashrightarrow{\pmb{\sf{ T_n = a  + (n - 1)d}}}}

{\dashrightarrow{\sf{ T_{15}= a  + (15 - 1) d}}}

{\dashrightarrow{\sf{ T_{15}= a  + (14) d}}}

{\dashrightarrow{\sf{ T_{15}= a  + 14 \times d}}}

{\dashrightarrow{\sf{ T_{15}= a  + 14d}}}

{\star{\underline{\boxed{\sf \pink{ T_{15}= a  + 14d}}}}}

Hence, the value of T₁₅ is a + 14d

\rule{190}1

Now, finding the difference between T₂₀ - T₁₅ :

{\dashrightarrow{\pmb{\sf{T_{20} -  T_{15}}}}}

{\dashrightarrow{\sf{(a + 19d) -  (a + 14d)}}}

{\dashrightarrow{\sf{a + 19d -  a  -  14d}}}

{\dashrightarrow{\sf{a - a + 19d -  14d}}}

{\dashrightarrow{\sf{0+ 19d -  14d}}}

{\dashrightarrow{\sf{19d -  14d}}}

{\dashrightarrow{\sf{5 \times  - 4}}}

{\dashrightarrow{\sf{ - 20}}}

{\star \: \underline{\boxed{\sf{\pink{T_{20} -  T_{15} =  - 20}}}}}

Hence, the value of T₂₀ - T₁₅ is -20.

\underline{\rule{220pt}{3.5pt}}

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