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juin [17]
4 years ago
12

Which term describes lines that meet at right angles?

Mathematics
1 answer:
charle [14.2K]4 years ago
4 0
D. Perpendicular lines is the relationship between two lines that meet at a right angle.
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What are the zeros of the quadratic function f(x) = 6x2 + 12x - 7?<br>​
mina [271]

Answer:

\frac{-6+\sqrt{78} }{6} and \frac{-6-\sqrt{78} }{6}

Step-by-step explanation:

We are asked that what are the zeros of the quadratic function f(x) = 6x² +12x -7.

So, we have to find the roots of the equation, f(x) = 6x² +12x -7 =0 ...... (1)

Since the quadratic function can not be factorized, so we have to apply Sridhar Acharya's formula.

This formula gives if, ax² +bx +c =0, the the two roots of the equation are

\frac{-b+\sqrt{b^{2}-4ac } }{2a} , \frac{-b-\sqrt{b^{2}-4ac } }{2a}

Therefore, in our case 'a' being 6, 'b' being 12 and 'c' being -7, the two roots of the equation (1) will be

\frac{-12+\sqrt{12^{2}-4*6*(-7) } }{2*6},  \frac{-12-\sqrt{12^{2}-4*6*(-7) } }{2*6}

= \frac{-12+\sqrt{312} }{12},\frac{-12-\sqrt{312} }{12}

=\frac{-6+\sqrt{78} }{6} and \frac{-6-\sqrt{78} }{6}

Hence, x= \frac{-6+\sqrt{78} }{6}

and x= \frac{-6-\sqrt{78} }{6}

(Answer)

9 0
4 years ago
Frankie buys six and twelve fifteenths pounds of cherries and four and four sixths pounds of grapes. How many pounds of fruit di
LUCKY_DIMON [66]

Answer:

Eleven and seven fifteenths pounds of fruit.

Step-by-step explanation:

6+4 = 10 pounds

12/15 + 4/6 = 1+7/15

Therefore, total pounds of fruit = 11 and 7/15

Hope this helped!

6 0
3 years ago
What has prompted you to borrow money from friends or family?
Tcecarenko [31]

Answer:

I needed money  to buy stuff at the mall with my friends because I didn't want to spend my own... :/ lol

6 0
3 years ago
Read 2 more answers
What is the equation of the line that passes through the point (-2,14) and is perpendicular to the line with the following equat
tino4ka555 [31]

keeping in mind that perpendicular lines have negative reciprocal slopes, let's check for the slope of the equation above

y=\stackrel{\stackrel{m}{\downarrow }}{-\cfrac{2}{5}}x-1\qquad \impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array} \\\\[-0.35em] ~\dotfill

\stackrel{~\hspace{5em}\textit{perpendicular lines have \underline{negative reciprocal} slopes}~\hspace{5em}} {\stackrel{slope}{\cfrac{-2}{5}} ~\hfill \stackrel{reciprocal}{\cfrac{5}{-2}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{5}{-2}\implies \cfrac{5}{2}}}

so we're really looking for the equation of a line whose slope is 5/2 and it passes through (-2 , 14)

(\stackrel{x_1}{-2}~,~\stackrel{y_1}{14})\hspace{10em} \stackrel{slope}{m} ~=~ \cfrac{5}{2} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{14}=\stackrel{m}{ \cfrac{5}{2}}(x-\stackrel{x_1}{(-2)}) \implies y -14= \cfrac{5}{2} (x +2) \\\\\\ y-14=\cfrac{5}{2}x+5\implies {\Large \begin{array}{llll} y=\cfrac{5}{2}x+19 \end{array}}

3 0
1 year ago
Please help im in 9th garde but this feels like 12th garde math soooo!
Stells [14]

Answer: don't know sorry

Step-by-step explanation:

4 0
2 years ago
Read 2 more answers
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