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Likurg_2 [28]
3 years ago
5

Identifying the values a, b, and c is the first step in using the Quadratic Formula to find solution(s) to a quadratic equation.

What are the values a, b, and c in the following quadratic equation? −6x2 = −9x + 7
Mathematics
2 answers:
lesya692 [45]3 years ago
3 0

The values of a, b, and c in given quadratic equation are:

a = 6 and b = -9 and c = 7

<em><u>Solution:</u></em>

Given quadratic equation is:

-6x^2 = -9x + 7

Let us first convert the given quadratic equation to standard form

The standard form is ax^2+bx+c=0 with a, b, and c being constants, or numerical coefficients, and x is an unknown variable

-6x^2 = -9x + 7\\\\-6x^2+9x-7=0\\\\6x^2-9x+7=0

Now we have to find the values of a, b, c

\text {For a quadratic equation } a x^{2}+b x+c=0, \text { where } a \neq 0\\\\x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}

Thus comparing 6x^2-9x+7=0 with standard form of quadratic equation a x^{2}+b x+c=0

a = 6

b = -9

c = 7

Thus values of a, b, and c are found

grin007 [14]3 years ago
3 0

Answer:

a=-6 b=-9 c=7

i just took this test and i got it right

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∆ ABC is similar to ∆DEF and their areas are respectively 64cm² and 121cm². If EF = 15.4cm then find BC.​
lyudmila [28]

{\large{\textsf{\textbf{\underline{\underline{Given :}}}}}}

★ ∆ ABC is similar to ∆DEF

★ Area of triangle ABC = 64cm²

★ Area of triangle DEF = 121cm²

★ Side EF = 15.4 cm

{\large{\textsf{\textbf{\underline{\underline{To \: Find :}}}}}}

★ Side BC

{\large{\textsf{\textbf{\underline{\underline{Solution :}}}}}}

Since, ∆ ABC is similar to ∆DEF

[ Whenever two traingles are similar, the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. ]

\therefore \tt \boxed{  \tt \dfrac{area( \triangle \: ABC )}{area( \triangle \: DEF)} =  { \bigg(\frac{BC}{EF} \bigg)}^{2}   }

❍ <u>Putting the</u><u> values</u>, [Given by the question]

• Area of triangle ABC = 64cm²

• Area of triangle DEF = 121cm²

• Side EF = 15.4 cm

\implies  \tt  \dfrac{64   \: {cm}^{2} }{12 \:  {cm}^{2} }  =  { \bigg( \dfrac{BC}{15.4 \: cm} \bigg) }^{2}

❍ <u>By solving we get,</u>

\implies  \tt    \sqrt{\dfrac{{64 \: cm}^{2} }{ 121 \: {cm}^{2} }}   =   \bigg( \dfrac{BC}{15.4 \: cm} \bigg)

\implies  \tt    \sqrt{\dfrac{{(8 \: cm)}^{2} }{  {(11 \: cm)}^{2} }}   =   \bigg( \dfrac{BC}{15.4 \: cm} \bigg)

\implies  \tt    \dfrac{8 \: cm}{11 \: cm}    =   \dfrac{BC}{15.4 \: cm}

\implies  \tt    \dfrac{8  \: cm \times 15.4 \: cm}{11 \: cm}    =   BC

\implies  \tt    \dfrac{123.2 }{11 } cm   =   BC

\implies  \tt   \purple{  11.2 \:  cm}   =   BC

<u>Hence, BC = 11.2 cm.</u>

{\large{\textsf{\textbf{\underline{\underline{Note :}}}}}}

★ Figure in attachment.

\rule{280pt}{2pt}

4 0
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An unknown number y is 20 more than an unknown number x. The number y is also x less than 2. The equations to find x and y are s
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Answer:

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Step-by-step explanation:

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11=x +20            -9 = x

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