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stepan [7]
3 years ago
5

MATH HELP ASAP!! MARKING BRAINLIEST

Mathematics
2 answers:
olga nikolaevna [1]3 years ago
7 0

Answer:

-1, -7, -13, -19

Step-by-step explanation:

-2(-3)-7 =6-7= -1

-2(0)-7 =0-7= -7

-2(3)-7 =-6-7= -13

-2(6)-7 =-12-7= -19

Tresset [83]3 years ago
3 0

Answer:

(-3,-1)(0,-7)(3,-13)(6,-19)

Step-by-step explanation:

Put the x number into the x in the equation to get your answer:

y=-2(-3)-7    y=-1

y=-2(0)-7     y=-7

y=-2(3)-7      y=-13

y=-2(6)-7     y=-19

(hope this helps :P)

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I don't get this question can someone help?
dedylja [7]

Answer:

I dont know

Step-by-step explanation:

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6 0
3 years ago
The pair (-1, 1) is a solution to the equation 2y - 3x = 5. Find another (x, y) pair that is a
morpeh [17]

Answer: (1,4)

Step-by-step explanation:

Given

Equation of line is 2y-3x=5

(-1,1) is the solution of the equation

Take x=1 and substitute it in the above equation

\Rightarrow 2y-3(1)=5\\\Rightarrow 2y=5+3\\\Rightarrow y=4

So, another pair of the solution is (1,4)

6 0
3 years ago
There are 12 more boys than girls at a school play. The ratio of boys to girls is 7:5. How many students were there altogether
taurus [48]

12 more boys than girls

ratio of boys to girls = 7:5

thus 2/12 is difference between boys and girls = 12

total students = 12 ÷ 2/12 = 72

7 0
3 years ago
The points A(1, 4), B(5,1) lie on a circle. The line segment AB is a chord. Find the equation of a diameter of the circle.
tangare [24]

Check the picture below.

well, we want only the equation of the diametrical line, now, the diameter can touch the chord at any several angles, as well at a right-angle.

bearing in mind that <u>perpendicular lines have negative reciprocal</u> slopes, hmm let's find firstly the slope of AB, and the negative reciprocal of that will be the slope of the diameter, that is passing through the midpoint of AB.

\bf A(\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\qquad B(\stackrel{x_2}{5}~,~\stackrel{y_2}{1}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{1}-\stackrel{y1}{4}}}{\underset{run} {\underset{x_2}{5}-\underset{x_1}{1}}}\implies \cfrac{-3}{4} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{slope of AB}}{-\cfrac{3}{4}}\qquad \qquad \qquad \stackrel{\textit{\underline{negative reciprocal} and slope of the diameter}}{\cfrac{4}{3}}

so, it passes through the midpoint of AB,

\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ A(\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\qquad B(\stackrel{x_2}{5}~,~\stackrel{y_2}{1}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{5+1}{2}~~,~~\cfrac{1+4}{2} \right)\implies \left(3~~,~~\cfrac{5}{2} \right)

so, we're really looking for the equation of a line whose slope is 4/3 and runs through (3 , 5/2)

\bf (\stackrel{x_1}{3}~,~\stackrel{y_1}{\frac{5}{2}}) \stackrel{slope}{m}\implies \cfrac{4}{3} \\\\\\ \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}{\cfrac{5}{2}}=\stackrel{m}{\cfrac{4}{3}}(x-\stackrel{x_1}{3})\implies y-\cfrac{5}{2}=\cfrac{4}{3}x-4 \\\\\\ y=\cfrac{4}{3}x-4+\cfrac{5}{2}\implies y=\cfrac{4}{3}x-\cfrac{3}{2}

4 0
3 years ago
Find all roots of the polynomial function F(x)=x^2-2x-24
liq [111]
f(x)=x^2-2x-24=x^2+4x-6x-24=x(x+4)-6(x+4)=\\\\=(x+4)(x-6)\\\\f(x)=0\ \ \ \Leftrightarrow\ \ \ (x+4)(x-6)=0\\\\.\ \ \ \ \ \ \ \ \ \ \  \ \ \Leftrightarrow\ \ \ x+4=0\ \ \ \ \ or\ \ \ \ \ x-6=0\\\\.\ \ \ \ \ \ \ \ \ \ \  \ \ \Leftrightarrow\ \ \ x=-4\ \ \ \ \ \ \ \ or\ \ \ \ \ x=6
4 0
3 years ago
Read 2 more answers
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