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2 years ago

Work out the value of angle x.Х62°

Mathematics

2 answers:

butalik [34]2 years ago

8 0

Answer:

because the first side equal for the second side this means that this triangle is isosceles

so 62+62=124

sum of the angles of any triangle =180

180°-124=56degree

kondaur [170]2 years ago

3 0

Answer:

62 grados x

Step-by-step explanation:

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andriy [413]

You can use factors to solve. Determine all the factor pairs of 24, find the two that are two numbers apart.
1, 24 X
2, 12 X
3, 8 X
4, 6 YES!

Algebraic way to solve using Quadratics:
l = 2 + w

A = lw

A = (2 + w)w Substitute (2 + w) for l

24 = (2 + w)w Substitute 24 in for the area

24 = 2w + w^2 Distribute

w^2 + 2w - 24 = 0 Set equal to 0 (put in standard form)

(w + 6) (w - 4) = 0 Factor

w + 6 = 0 and w - 4 = 0 Set each factor equal to 0.

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3 years ago

A university's freshman class has 5000 students. 50% of those students are majoring in Computer Science. How many students in th

Mazyrski [523]

2500 students are computer science majors

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2 years ago

An ant weighs 2 x 10^-3 grams, and a lion weighs 1 x 10^5. The lion's weight is 5 x 10^x times greater than the ant's weight. Wh

julsineya [31]

Answer:

x=7

Step-by-step explanation:

Information provided

Weight of ant= $2*10^{-3} grams$

Weight of lion= $1*10^{5}$ but units are not provided, assuming this unit is also grams

How many times lion's weight exceed an ant

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= $5*10^{7}$

Therefore, the value of x=7 if the unit for weight of lion is assumed to be same as unit for weight of an ant

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2 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)

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