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2 years ago

Please help on both! And please put answer and not a website!

Mathematics

1 answer:

marissa [1.9K]2 years ago

3 0

Answer:

1: 55 2: 40

Step-by-step explanation:

For the first problem, by the triangle angle theorm, the three angles in a triangle add up to 180, so

90+35+x=180

125 + x = 180

x = 55

Now, the angle you just found (x = 55) and the exterior angle are supplementary so they also add up to 180

55 + y = 180

y = 125

For the second problem, the two angles in a triangle non-adjacent to the exterior angle are equal to the exterior angle, so

40 + 2x = 3x

x = 40

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I dont understand please help Which expression is equivalent to 8-(6r+2) A -6r+6 B 2r+2 C 6r+10 D -6r+10

Goryan [66]

Answer:

-6r+6

Step-by-step explanation:

Given data

We are given the expression

8-(6r+2)

let us expand it by opening the bracket

=8-6r-2

collect like terms

=8-2-6r

=6-6r

rearrange

=-6r+6

Hence option A is correct

8 0

2 years ago

Determine if the following system of equations has no solutions 4x+3y= -8 -8x-6y= 16

cupoosta [38]

$\begin{cases} 4x+3y=-8\\\\ -8x-6y=16 \end{cases}~\hspace{10em} \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array} \\\\[-0.35em] ~\dotfill$

$4x+3y=-8\implies 3y=-4x-8\implies y=\cfrac{-4x-8}{3}\implies y=\stackrel{\stackrel{m}{\downarrow }}{-\cfrac{4}{3}} x-\cfrac{8}{3} \\\\[-0.35em] ~\dotfill\\\\ -8x-6y=16\implies -6y=8x+16\implies y=\cfrac{8x+16}{-6} \\\\\\ y=\cfrac{8}{-6}x+\cfrac{16}{-6}\implies y=\stackrel{\stackrel{m}{\downarrow }}{-\cfrac{4}{3}} x-\cfrac{8}{3}$

one simple way to tell if both equations do ever meet or have a solution is by checking their slope, notice in this case the slopes are the same for both, meaning the lines are parallel lines, however, notice both equations are really the same, namely the 2nd equation is really the 1st one in disguise.

since both equations are equal, their graph will be of one line pancaked on top of the other, and the solutions is where they meet, hell, they meet everywhere since one is on top of the other, so infinitely many solutions.

3 0

1 year ago

Find (a) arc length and (b) Area of a sector.

barxatty [35]

Answer:

a) 29.45 cm (2 dp)

b) 220.89 cm² (2 dp)

Step-by-step explanation:

<u>Formula</u>

$\textsf{Arc length}=r \theta$

$\textsf{Area of a sector}=\dfrac{1}{2}r^2 \theta$

$\quad \textsf{(where r is the radius and}\:\theta\:{\textsf{is the angle in radians)}$

<u>Calculation</u>

Given:

$\theta=\dfrac{5 \pi}{8}$
r = 15 cm

$\begin{aligned}\implies \textsf{Arc length} & =r \theta\\& = 15\left(\dfrac{5 \pi}{8}\right)\\& = \dfrac{75}{8} \pi \\& = 29.45\: \sf cm\:(2\:dp)\end{aligned}$

$\begin{aligned} \implies \textsf{Area of a sector}& =\dfrac{1}{2}r^2 \theta\\\\ & = \dfrac{1}{2}(15^2) \left(\dfrac{5 \pi}{8}\right)\\\\& = \dfrac{225}{2}\left(\dfrac{5 \pi}{8}\right)\\\\ & = \dfrac{1125}{16} \pi \\\\& = 220.89 \: \sf cm^2\:(2\:dp)\end{aligned}$