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PolarNik [594]
3 years ago
6

Can you construct a triangle with length of 11, 17, 29?

Mathematics
1 answer:
KATRIN_1 [288]3 years ago
8 0
No, because 11+17=28\not>29 and the sum of two sides must be greater than the third side.
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Use this figure to find the value of ∠FOD. <br><br> A. 27°<br> B. 63°<br> C. 46°<br> D. 78°
BartSMP [9]

m<FOD = 90 -12 -15

m<FOD = 63

answer is B. 63

hope it helps

7 0
4 years ago
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Subract 7 1/5 subract 6 2/5
mina [271]

Answer:

1.42 is the answer

Step-by-step explanation:

3 0
3 years ago
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Use models to solve each inequality. Write the answer in set builder notation. 4-7?!
butalik [34]

Answer:


Step-by-step explanation:

4) Given that 3x<15

Divide by +3 without affecting inequality

x<5

Hence answer is (-∞,5)

5) -6x<18

Add 6x to both sides

0<6x+18

Subtract 18

-18<6x

Divide by +6

-3<x

Hence answer is (-3,∞)

6) 2x+6>x-7

Subtract x-6 from both the sides

x >-13

So answer is (-13,∞)

7) -4x+8≥14

Add 4x-14 to  both the sides

-6≥4x

x≥-3/2

Hence answer is [-3/2,∞)

8 0
3 years ago
Given h(x) = -x - 5, find h(0)​
fenix001 [56]

Answer:

-5

Step-by-step explanation:

you plug in zero for x and then just solve.

h(0)=-0-5

=-5

5 0
3 years ago
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I NEED HELP FAST PLEASE!!!!!
raketka [301]

Answer:

Excluded values  -2,0,2  These are non removable since we cannot cancel them in the numerator

(x+1)

---------

x(x+2)  

Step-by-step explanation:

-2                  x-1

----------- + ---------

x^2 -4        x^2 -2x

First factor the denominators

-2                  x-1

----------- + ---------

x^2 -4        x^2 -2x

x^2 -4 is the difference of squares

x^2 -4 = (x-2) (x+2)

x^2 -2x has an x in common

x(x-2)

-2                   x-1

----------- +    ---------

(x-2)(x+2)        x(x -2)

The values that are excluded are the values that make the denominators zero

(x-2) (x+2) = 0     x=2  x=-2

x(x-2) =0  x=0  x=2

Excluded values  -2,0,2  These are non removable since we cannot cancel them in the numerator

We need to get a common denominator of (x-2)(x+2)x

Multiply the first term by x/x and the second term by (x+2)/(x+2)

-2 x                 (x-1) (x+2)

----------- +    ----------------

x(x-2)(x+2)        x(x -2)(x+2)

Distribute

(x+2)(x-1) = x^2 -x+2x-2 = x^2+x-2

-2 x                x^2+x-2

----------- +    ----------------

x(x-2)(x+2)        x(x -2)(x+2)

Combine like terms, since the denominators are the same

-2 x  + x^2+x-2

---------------------------

x(x-2)(x+2)    

 x^2-x-2

---------------------------

x(x-2)(x+2)      

Factor the numerator

 x^2-x-2 = (x-2) (x+1)

(x-2) (x+1)

---------------------------

x(x-2)(x+2)  

Cancel like terms

(x+1)

---------------------------

x(x+2)  

The excluded values are found at the beginning, not now.  We canceled some of them in our simplified expression.  They are still excluded values

3 0
3 years ago
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