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Alexxandr [17]
3 years ago
15

What is the product of 2x+3 and 4x^2-5x+6

Mathematics
1 answer:
solniwko [45]3 years ago
7 0
(2x+3)(4x^2 - 5x + 6) = 8x^3 - 6x^2 - 3x + 18
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Find all of the zeros
noname [10]

Answer:

{-3, -1, 1}

Step-by-step explanation:

Zeros refer to x-intercepts. X-intercepts are x values when y = 0. You can tell where they are by looking at where the line passes the x-axis.

Therefore, {-3, -1, 1} are the zeros.

3 0
4 years ago
The sum of two numbers a and b is equal to c. The number line below shows the values of a and c. Shat is the value of b?​
Volgvan

Answer:

HIS SGSEIJGR

Step-by-step explanation:

3 0
3 years ago
Find the perimeter of the figure to the nearest hundredth.​
WITCHER [35]
Perimeter: 36.84 feet

Let’s start with the easy part...all the rectangle pieces 6 x 3 = 18 feet

We know from subtracting the rectangle length (6ft) from the total length (12 ft) that the diameter of the circle is 6ft.

Circumference = diameter x pi
= 6 x 3.14 = 18.84

We have the circle split into 2 halves, but the circumference remains the same

18 + 18.84 = 36.84
6 0
3 years ago
Can you help me find the value of x and y. I am very confused.
Monica [59]
Hello,
Let's assume top left  corner: A
top right corner : B
Bottom right corner: C
Bottom left corner :D

M= middle of [CD]

ABM is a triangle rectangular isocel:

(3√2)²+(3√2)²=y²
==>y²=2*9*2
==>y²=36
==>y=6

The triangle BCM is rectangular with MB=3√2, MC=y/2=3
x²+3²=(3√2)²
==>x²=9*2-9
==>x²=9
==>x=3

8 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
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