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Alekssandra [29.7K]
3 years ago
6

Bill began his diet when he weighed 268 pounds. After 4 weeks he weighed 250 pounds. Write an equation in slope-intercept form o

f the line if w represents weeks and p represents pounds
Mathematics
1 answer:
Alika [10]3 years ago
7 0

An equation in slope-intercept form of the line

p=-\frac{9}{2}w+268\\

Given :

Bill began his diet when he weighed 268 pounds.

After 4 weeks he weighed 250 pounds

Lets 'w' represents weeks and 'p' represents pounds

Using the given information ,we find two points (w,p)

Bill began his diet when he weighed 268 pounds.

when w=0 , weight in pounds =268

(0,268)

After 4 weeks he weighed 250 pounds, so point is (4,250)

Use two points to write the equation of line

Slope intercept form is y=mx+b

where m is the slope and b is the y intercept

slope formula using 2 points (0,268) and (4,250)

m=\frac{y_2-y_1}{x_2-x_1} =\frac{250-268}{4-0} =-\frac{9}{2}

now use slope m and any one point (0,268) to write equation

y=mx+b\\y=-\frac{9}{2}x+b\\(0,268) where x=0  and y=268\\268=-\frac{9}{2}(0)+b\\\\b=268

we know that w represents weeks and p represents pounds

The equation of the line in slope intercept form is

p=-\frac{9}{2}w+268\\

Learn more :  brainly.com/question/986503

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Harry is trying to solve the equation 0 = 2x2 − x − 6 using the quadratic formula. He has made an error in one of the steps belo
bija089 [108]
<h3><u>Given</u> - </h3>

➙ a quadratic equation in which Harry lagged due to an error made by him, 2x² - x - 6= 0

<h3><u>To solve</u> - </h3>

➙ the given quadratic equation.

<h3><u>Concept applied</u> - </h3>

➙ We will apply the quadratic formula as given in the question. So, let's study about quadratic equation first because we are supposed to apply the formula in equation.

What is quadratic equation?

➙ A quadratic equation in the variable x is an equation of the form ax² + bx + c = 0, where a, b, c are real numbers, a ≠ 0.

Now, what is quadratic formula?

➙The roots of a quadratic equation ax + bx + c = 0 are given by \sf{\:\frac{-b \pm\: \sqrt {b ^ 2 - 4ac}}{2a}} provided b - 4ac ≥ 0.

<h3><u>Solution</u> - </h3>

here as per the given quadratic equation,

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putting in the formula,

\implies\sf{x=\frac{-(-1) \pm\: \sqrt {(-1)^2 - 4(2)(-6)}}{2(2)}}

\implies\sf{x=\frac{1 \pm\: \sqrt {1+48}}{4}}

\implies\sf{x=\frac{1 \pm\: \sqrt {49}}{4}}

\implies\sf{x=\frac{1 \pm\: 7}{4}}

Solving one by one,

\implies\sf{x=\frac{1 + \: 7}{4}}

\implies\sf{x=\frac{8}{4}}

\implies{\boxed{\bf{x=2}}}

________________

\implies\sf{x=\frac{1 - \: 7}{4}}

\implies\sf{x=\frac{-6}{4}}

\implies{\boxed{\bf{x=\frac{-3}{2}}}}

________________________________

<em><u>Note</u> - Hey dear user!! You haven't provided the solution which was solved by Harry (A.T.Q). Please go through the solution as it will help you to find the error done by Harry.</em>

<em>________________________________</em>

Hope it helps!! (:

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Answer:

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Hope this helped.

4 0
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