i need some answers checked

8 months ago@Vocaloid my choice was the top right one

8 months agoThe leg and hypotenuse cannot both be 6. Remember what we said earlier: the leg of a 45-45-90 triangle is the hypotenuse divided by sqrt(2)

8 months agoso it was the top left one?

8 months agoHypotenuse ~divided~ by sqrt(2) not multiplied.

8 months agoso the bottom right one, as 3 is a multiple of 6?

8 months agoYes 6/sqrt(2) = 3sqrt(2)

8 months agoi got A on this one

8 months agoNope First calculate the side length of the square using perimeter = 4s Then multiply by sqrt(2) because hypotenuse = leg * sqrt(2)

8 months agoIf the perimeter of a square is 10, then side length =?

8 months agoPerimeter = 4* side length Solve for side length

8 months agoso if the perimeter is 4, wouldnt that make it 1.8?

8 months agoThe perimeter is not 4 Please re-read the original problem

8 months agooh, right...

8 months agoA square has four equal sides right? So the side length = perimeter/4

8 months ago10/4=2.5

8 months agoGood now multiply that by sqrt(2)

8 months ago6.25

8 months agoCheck your calculations again Remember order of operations. Take the square root of 2 first and then multiply by 6

8 months ago*multiply by 2.5 not 6

8 months ago24?

8 months agohang on, i might've did it wrong again

8 months agoPress the square root button, then 2 Press the enter button Then multiply by 2.5

8 months agoso 4, x 2.5 = 10

8 months agoSquare root is not the same thing as squaring So the square root of 2 is not 4

8 months agoTry this 2^(0.5) * 2.5

8 months ago3.53

8 months agoGood 3.5 is the only solution close enough so that's our answer

8 months agowhat i got on this, was 16.

8 months agoHm not quite, 16 would be the value of x not y To get y you just need to multiply 16 by sqrt(3) since y is the side across the 60 degree side

8 months agoso D?

8 months agoGood

8 months agoi chose the last one, as an isosceles is suppose to have two congruent sides, right?

8 months agoGood

8 months agoi chose C for this one

8 months agoHm not quite, 8 would be for a 30-60-90 triangle For a 45-45-90 triangle hypotenuse = leg * sqrt(2) =?

8 months ago8?

8 months agoor at least, 16?

8 months agoThe leg is 4 right? So sqrt(2) * 4 =?

8 months ago5.6

8 months agoso b

8 months agoGood

8 months agofalse

8 months agoGood

8 months agoA?

8 months agoThe height of the wall is basically one leg of a right triangle so by definition it cannot be longer than the hypotenuse For a 45-45-90 triangle Leg = hypotenuse / sqrt(2) =?

8 months agoI mean, even without doing any calculations there's only one choice less than the hypotenuse 13

8 months ago9.2?

8 months agoYup good

8 months agothe top right one?

8 months agoGood

8 months agoXZ

8 months agoHm that's a very good guess but XZ is the hypotenuse which is not considered a leg So between the other two sides which one do you think is longer?

8 months agoXY

8 months agoGood so xy is your sol 'n

8 months agomy choice is C

8 months agoHm not quite You're just gonna count the number of horizontal spaces between M and N

8 months agoin other words just subtract the x-coordinates of M and N

8 months ago* M and F I mean

8 months agoso you have a right triangle and you are given the hypotenuse and one leg so try plugging them into a^2 + b^2 = c^2 and solving for the missing leg

8 months ago1300

8 months ago36.05

8 months agohm, not quite, remember we have the hypotenuse (30) and one leg (20) so 30^2 = a^2 + 20^2 solve for a

8 months ago500

8 months agoalmost, that's a^2 so you'd just have to take the square root of 500

8 months ago22.3

8 months agoalmost, the square root of 500 is 22.360 something so it rounds up to 22.4 not 22.3

8 months agoso thats my answer?

8 months agoyes

8 months agothis is true

8 months agohm, be careful, it's the sum of the squares of the lengths, not just the lengths so false

8 months agothe top right one

8 months agohm, not quite, a good thing to remember is that the hypotenuse is the biggest side so it has to be bigger than 10 a^2 + b^2 = c^2 plug in your legs (10 and 4) and solve for c

8 months agoso the bottom left one? that's what i got from it

8 months agoremember to square the sides before adding them 10^2 + 4^2 = ?

8 months ago116

8 months agogood and what's the square root of that?

8 months agoi disconnected, sorry

8 months ago10.7

8 months agogood so 10.77 is the closest and thus your sol'n

8 months ago24?

8 months agogood

8 months agowhen you have a diagonal through a square you get two 45-45-90 triangles as usual leg = hypotenuse/sqrt(2) = ?

8 months ago3.5?

8 months ago12.25 i meant

8 months agohm not quite what is 3.5 divided by sqrt(2)? remember order of operations

8 months agoI don't really know how to explain this any better haha you just need to type 3.5/√(2)

8 months ago2.47

8 months agogood but they want the nearest tenth so 2.5

8 months ago7?

8 months agohm, not 7. remember you cannot just add the side lengths together, you have to use the formula a^2 + b^2 = c^2 to solve for c

8 months agowhen you sketched the triangle you got side lengths 3 and 4 right? plug those in as your a and b values then solve for c.

8 months ago5

8 months agogood so 5 = your sol'n

8 months ago3?

8 months agonot quite what are the x-coordinates of B and Z?

8 months ago-9 and -5

8 months agogood so what's the difference of -9 and -5?

8 months ago-4 apart?

8 months agogood but since it's distance we take the absolute value of that to get 4 = your sol'n

8 months agoC?

8 months agonot quite you are given the hypotenuse and one leg a^2 + b^2 = c^2 solve for the missing leg remember that c is the hypotenuse and (a and b) are the legs

8 months agothe hypotenuse is 65 and one of the legs is 33 right? therefore 33^2 + a^2 = 65^2 so solve for (a)

8 months ago49?

8 months ago33^2 + a^2 = 65^2 subtract 33^2 from both sides to get a^2 = 65^2 - 33^2 therefore a = ?

8 months ago5134

8 months agothat's a^2 so if a^2 = 5134 a = ?

8 months ago71.6

8 months agowait a minute how did you get 5134

8 months agowhat's 65^2 - 33^2?

8 months ago65^2 = ? 33^2 = ? calculate the two values separately then subtract them

8 months ago3136

8 months ago56

8 months agogood so 56 = your sol'n

8 months ago17?

8 months agogood

8 months agotop right?

8 months agohm not quite you have the opposite and adjacent, what trig ratio is this?

8 months agouh...

8 months agodang, i just forgot trig...

8 months ago|dw:1529378231682:dw|

8 months agoas I stated before you have the adjacent and opposite sides which trig ratio is appropriate? check the diagram

8 months agowhich trig ratio involves the opposite and adjacent sides? check the red text

8 months ago|dw:1529379233861:dw|

8 months agoso you'd use tangent, right? so you set up tan(theta) = opposite/adjacent plug in your values & solve for the missing side.

8 months agoi got top left, for some reason, i dont think its right...

8 months agowhich side of the triangle is opposite to the angle? and which is the adjacent side?

8 months agothe dotted one is adjacent, right?

8 months agothe dotted side is the hypotenuse and thus is not included in this calculation.

8 months agosorry, i'm not working at full capacity...

8 months agoi'll keep trying though

8 months agoso there are only two other sides of the triangle are 21 and x notice how the x side is the farthest across from the angle, so that's our opposite side the 21 is next to the angle so that's our adjacent side therefore tan(36) = x/21 solve for x

8 months agoi am so sorry, i passed out...

8 months ago@Vocaloid let us continue where we left off

8 months agosure solving for x gives us tan(36) * 21 which you can then just plug into a calculator to get the sol'n

8 months ago162

8 months agocan you try putting in tan(36) then multiply the result by 21 ?

8 months agostill 162

8 months agooh wait your calculator is probably in radians mode can you try changing it to degrees mode?

8 months ago15.25?

8 months agogood it rounds up to 15.26 actually so 15.26 is your sol'n

8 months ago@Vocaloid

8 months agofrom the perspective of angle A opposite/adjacent = ? remember that the hypotenuse is not included

8 months ago|dw:1529431215885:dw|

8 months ago^ try to apply this logic to your triangle to figure out what the adjacent and opposite sides are remember that the opposite side is the furthest away from the angle, and the adjacent side is closest w/o being the hypotenuse

8 months agostill there? try to see which side is farthest away from angle A and which one is the closest to angle A

8 months agoyeah, thinking

8 months agoi somehow got the bottom left one

8 months agothat's the adjacent over the opposite tan is opposite over adjacent

8 months agothe side farthest away from A is the 5 side the side next to A is the 12 side so opposite/adjacent = ?

8 months ago\(\color{#0cbb34}{\text{Originally Posted by}}\) @Vocaloid ^ try to apply this logic to your triangle to figure out what the adjacent and opposite sides are remember that the opposite side is the furthest away from the angle, and the adjacent side is closest w/o being the hypotenuse \(\color{#0cbb34}{\text{End of Quote}}\)

8 months agotherefore the opposite side is 5 and the adjacent side is 12 making tan = opposite/adjacent = 5/12 = your solution gtg

8 months agosorry , see you later

8 months ago