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Gwar [14]
3 years ago
6

Just need help with the blank ones please ! Will give brainiest .

Mathematics
1 answer:
weeeeeb [17]3 years ago
5 0
Answers:
11. B) Acute
12. D) BA = QP
13. C) 36
15. C) BC = ST
16. Hypontenuse is roughly 23.6 millimeters
17. C) 17 ft
------------------------------------------------
Work Shown

Problem 11)
T+U+V = 180
38+U+72 = 180
38+72+U = 180
110+U = 180
110+U-110 = 180-110
U = 70
All three angles (38, 72, 70) are less than 90 degrees. This triangle is acute
-----------------
Problem 12)
If triangle ABC = triangle PQR, then the corresponding parts must be congruent. So,
AB = PQ
BC = QR
AC = PR
angle A = angle P
angle B = angle Q
angle C = angle R
The only choice that fits with the equations mentioned above is choice D, which is why it's the answer
Keep in mind BA is the same as AB. The order doesn't matter
-----------------
Problem 13)
FP is twice as long as PY
FP = 2*PY
FY = FP+PY
FY = 2*PY+PY
FY = 3*PY
Because FP = 24, we can find PY
FP = 2*PY
24 = 2*PY
PY = 12
Which is then used to find FY
FY = 3*PY
FY = 3*12
FY = 36
-----------------
Problem 15)
HL stands for "hypotenuse leg". We already have a pair of congruent legs which is RS = AB, given by the tick marks. We just need a pair of hypotenuses. In this case, BC and ST are the hypotenuse of the triangles. So if we knew BC = ST, then we'd have enough for HL to be used. 
-----------------
Problem 16)
a^2 + b^2 = c^2
14^2 + 19^2 = c^2
557 = c^2
c^2 = 557
c = sqrt(557) <<--- "sqrt" stands for "square root"
c = 23.60084
c = 23.6
The hypotenuse is roughly 23.6 millimeters long
-----------------
Problem 17)
Let a,b,c be the sides of the triangle. The given sides are a=6 and b=17. The unknown side is c. We don't know the exact value of c, but we can figure out how small and how large it can get.
It turns out that c is restricted through this compound inequality
b-a < c < b+a
Plug in the given values and simplify
b-a < c < b+a
17-6 < c < 17+6
11 < c < 23
So the third side can be between 11 and 23, but not equal to either endpoint. The only choice that fits this inequality is 17, so that's why the answer is C.
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Simon has more money than Kande. if Simon gave Kande K20, they would have the same amount. While if Kande gave Simon $22, Simon
motikmotik

Given parameters;

  Let us solve this problem step by step;

Let us represent Simon's money by S

Kande's money by K

  • Simon has more money than Kande

               S > K

  •  if Simon gave Kande K20, they would have the same amount;

 if Simon gives $20, his money will be  S - 20 lesser;

      When Kande receives $20, his money will increase to K + 20

                     S - 20  = K + 20   ------ (i)

  • While if Kande gave Simon $22, Simon would then have twice as much as Kande;

          if Kande gave Simon $22, his money will be K - 22

    Simon's money, S + 22;

                  S + 22  = 2(K - 22)    ------ (ii)

Now we have set up two equations, let us solve;

         S - 20  = K + 20  ---- i

         S + 22  = 2(K - 22)  ;       S + 22  = 2K - 44  ---- ii

So,      S - 20  = K + 20

          S + 22  = 2K - 44

subtract both equations;

               -20 - 22  = (k -2k)  + 64

                   -42  = -k + 64

                       k  = 106

Using equation i, let us find S;

            S - 20 = K + 20

             S - 20  = 106 + 20

              S = 106 + 20 + 20  = 146

Therefore, Kande has $106 and Simon has $146

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