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3 years ago

Please help!!

Mathematics

1 answer:

Soloha48 [4]3 years ago

3 0

Step-by-step explanation:

Given points are:

$(-1,\:3)=(x_1,\: y_1) \:\&\:(1, \:7)=(x_2,\:y_2)$

Equation of line in two point form is given as:

$\frac{y-y_1 }{y_1 - y_2} = \frac{x-x_1 }{x_1 - x_2} \\ \\ \therefore \: \frac{y - 3}{3 - 7} = \frac{x - ( - 1)}{ - 1 - 1} \\ \\ \therefore \: \frac{y - 3}{ - 4} = \frac{x + 1}{ - 2} \\ \\ \therefore \: \frac{y - 3}{ 2} = \frac{x + 1}{ 1} \\ \\ \huge \purple{ \boxed{\therefore \:y - 3 = 2(x + 1)}} \\ this \: is \: the \: equation \: of \: line \: in \: \\ point - slope \: form.\\simplifying \: it \: further \: we \: find: \\ y - 3 = 2x + 2 \\ \\ \therefore \:y = 2x + 2 + 3 \\ \\ \huge \red{ \boxed{\therefore \:y = 2x + 5}} \\ this \: is \: the \: equation \: of \: line \: in \: \\ slope - intercept \: form.$

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Write down the quadratic equation whose roots are $x = -7$ and $x = 1,$ and the coefficient of $x^2$ is 1. Enter your answer in

pav-90 [236]

<h2>Steps:</h2>

So firstly, since we know that the coefficient of x² is 1, this means that this is our base equation:

y = x² + bx + c

Now, since we know that the roots are -7 and 1, set y = 0 and set x = -7 and 1 and simplify:

$0=(-7)^2+b(-7)+c\\0=49-7b+c\\-49=-7b+c\\\\0=1^2+b(1)+c\\0=1+b+c\\-1=b+c\\\\-49=-7b+c\\-1=b+c$

Now with this, we can set up a system of equations to solve for b and c. For this, I will be using the elimination method. For this, subtract the 2 equations:

$\begin{alignedat}{2}-49&=-7b+c\\-(-1&=b+c)\\-48&=-8b\end{alignedat}$

Now that the c variable has been eliminated we can solve for b. For this, divide both sides by -8 and your first part of your answer is b = 6.

Now that we know the value of b, plug it into either equation to solve for c:

$-49=-7(6)+c\\-49=-42+c\\-7=c\\\\-1=6+c\\-7=c$

<h2>Answer:</h2>

<u>Putting it together, your final answer is x² + 6x - 7 = 0.</u>

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3 years ago

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3 years ago

What is the answer to (4.575 ÷ 275) × 2

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Answer:

0.0334

Step-by-step explanation:

(4.575 ÷ 275) × 2

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Answer:

$\fbox {Trapezoid ABCD and Trapezoid EFGH are congruent.}$

Step-by-step explanation:

<u>Finding corresponding side lengths</u> :

<u>Trapezium ABCD</u>

AB = √(-2)² + (-1)² = √5
BC = √(-1)² + (2)² = √5
CD = √(2)² + (2)² = 2√2
DA = √(-1)² + (5)² = √26

<u>Trapezium EFGH</u>

EF = √(2)² + (1)² = √5
FG = √(1)² + (2)² = √5
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As the corresponding side lengths are equal, we can conclude that the trapezoids are congruent.

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