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2 years ago

Evaluate 3x² - 1 when x = 2. A. 11 B. 35 C. 3 D. 5

Mathematics

1 answer:

Alexus [3.1K]2 years ago

4 0

Answer:

Step-by-step explanation:

All you need to do is substitute!

(3x2)^2-1

6^2-1

36-1

= 35

Hope this helps!!

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ILL GIVE BRAINLIEST!!

densk [106]

Answer:

P(3 , 0)

Step-by-step explanation:

(x₁, y₁) = A(-3,-3) ;(x₂ , y₂) B(5, 1)

m : n = 3 : 1

$P \left( \dfrac{mx_{2}+nx_1}{m+n},\dfrac{my_{2}+ny_{1}}{m+n} \right)\\\\P \left (\dfrac{5*3+(-3)*1}{3+1},\dfrac{3*1+1*(-3)}{3+1} \right)\\\\\\=P \left(\dfrac{15-3}{4},\dfrac{3-3}{4} \right)\\\\\\= P \left(\dfrac{12}{4},\dfrac{0}{4} \right)\\\\\\= P\left(3 , 0 \right)$

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3 years ago

Sarah is 13 years older than Mike. The sum of their ages is 37. How old is Sarah and how old is Mike? Enter your answer as on or

Blizzard [7]

Answer:

(24,13)

Step-by-step explanation:

24 is Sarah's age.

13 is Mike's age.

System:

$\left \{24+13=37$

I hope this helped!

3 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)

$\bf (\stackrel{x_1}{3}~,~\stackrel{y_1}{\frac{5}{2}}) \stackrel{slope}{m}\implies \cfrac{4}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{\cfrac{5}{2}}=\stackrel{m}{\cfrac{4}{3}}(x-\stackrel{x_1}{3})\implies y-\cfrac{5}{2}=\cfrac{4}{3}x-4 \\\\\\ y=\cfrac{4}{3}x-4+\cfrac{5}{2}\implies y=\cfrac{4}{3}x-\cfrac{3}{2}$

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3 years ago

What are the answer to these 5 questions this worth 20 points btw

Sunny_sXe [5.5K]

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3 years ago

How do i add 21p to 35q

mariarad [96]

I don't think you can because the variables are different. You can only add the same variables.

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3 years ago

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