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gogolik [260]
3 years ago
13

Find the value of k please! (There is a value and it’s not 14)

Mathematics
2 answers:
skelet666 [1.2K]3 years ago
8 0

Answer:

K=0

Step-by-step explanation:

14xk

k=0

14x0=10

Misha Larkins [42]3 years ago
5 0

Answer: k = 0

Step-by-step explanation:

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Can someone help me please ? Thanks!
notka56 [123]
Use pythagorean theorem for all of them
a. 3, 4, 5
3^2+4^2?5^2
9+16?25
25=25
right 

b. 5, 6, 7
5^2+6^2=7^2
25+36=49
61>49
acute

c. 64+81=144
145>144
acute

hope this helps!
6 0
3 years ago
Read 2 more answers
Tutorial Exercise A boat leaves a dock at 2:00 PM and travels due south at a speed of 20 km/h. Another boat has been heading due
ANTONII [103]

Answer:

Both the boats will closet together at 2:21:36 pm.

Step-by-step explanation:

Given that - At 2 pm boat 1 leaves dock and heads south and boat 2 heads east towards the dock. Assume the dock is at origin (0,0).

Speed of boat 1 is 20 km/h so the position of boat 1 at any time (0,-20t),

        Formula :   d=v*t

at 2 pm boat 2 was 15 km due west of the dock because it took the boat 1 hour to reach there at 15 km/h, so the position of boat 2 at that time was (-15,0)

the position of boat 2 is changing towards east, so the position of boat 2 at any time (-15+15t,0)

      Formula : D=\sqrt{(x2-x1)^2+(y2-y1)^2}

⇒                     D = \sqrt{20^2t^2+15(t-1)^2}

Now let           F(t) = D^2(t)

                ∵    F'(t) = 800t + 450(t-1) =  1250t -450\\F'(t) =0

⇒                     t= 450/1250

⇒                     t= .36 hours

⇒                       = 21 min 36 sec

Since F"(t)=0,

∴ This time gives us a minimum.

Thus, The two boats will closet together at 2:21:36 pm.

3 0
3 years ago
QUICKLY HELP!
Leviafan [203]

a+(−5)=−12

a−5=−12

a = -12 + 5

a = -7

answer is b. a=−7

7 0
3 years ago
Read 2 more answers
Is negitive 6 rational
n200080 [17]
-6 is a rational I believe
5 0
3 years ago
what is the solution to set to the inequality (4x-3)(2x-1)_>0 a) {x|x<_3 or x_>1} b) {x|x<_2 or x_> 4/3} c) {x|x&
Vinil7 [7]
       (4x - 3)(2x - 1) ≥ 0
4x - 3 ≥ 0    or    2x - 1 ≤ 0
    + 3  + 3                + 1 + 1
     4x ≥ 3                 2x ≤ 1
      4     4                  2    2
       x ≥ ³/₄                  x ≤ ¹/₂
        x ∈ [-∞, ¹/₂] ∧ [³/₄, ∞]

The answer is C.
6 0
2 years ago
Read 2 more answers
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