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never [62]
3 years ago
13

convert the mixed numbers to improper fractions. this will get you one step closer to finding the unit rate

Mathematics
1 answer:
Kaylis [27]3 years ago
8 0
It means for example instead of writing 2 2/5 You would write 12/5
You can find out an improper fraction by simply multiplying the denominator by the front number and adding the answer to you numerator
So, 2x5= 10. 10+2= 12. So 12/5
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Look at the triangles shown below.
Sophie [7]

Answer:

Variant c

Step-by-step explanation:

c²=4*9

c=6

You can apply phyphagor

b²=6²+9²=36+81=117

3 \sqrt{13}

4 0
3 years ago
Which equation is equivalent to the equation 6x+9=12? A: x + 9 = 6 B: 2 x + 3 = 4 C: 3 x + 9 = 6 D: 6 x + 12 = 9
Pachacha [2.7K]

Answer: B

Step-by-step explanation:

3 0
2 years ago
Find the value of x in kite EFGH<br>​
kolbaska11 [484]

Answer:

x = 105

Step-by-step explanation:

the sum of all interior angles of any four-sided figure has 360 degrees

∡F and ∡H are equal

add all four expressions and set them equal to 360

x + (x + 5) +( x + 5) + (140 - x) = 360

combine like terms

2x + 150 = 360

2x = 210

x = 105

3 0
3 years ago
Read 2 more answers
Hey guys<br>im new here<br>please solve this for me with steps!<br>ill mark as the best answer​
Vinil7 [7]

Answer:

The factors of  2(x+y)^2-9(x+y)-5 is ((x+y)-5)(2x+2y+1)

Step-by-step explanation:

Given polynomial

=>2(x+y)^2-9(x+y)-5

To Find:

The factors of the polynomial =?

Solution:

Lets assume  k = (x+y)

Then 2(x+y)^2-9(x+y)-5 can be written as 2k^2-9k-5

Now by using quadratic formula

k =\frac{-b\pm\sqrt{(b^2-4ac}}{2a}

where

a= 2

b= -9

c= -5

Substituting the values, we get

k =\frac{-b\pm\sqrt{(b^2-4ac)}}{2a}

k =\frac{-(-9) \pm \sqrt{((-9)^2-4(2)(-5)}}{2(2))}

k =\frac{-(-9) \pm \sqrt{(81+40)}}{4}

k =\frac{-(-9) \pm \sqrt{(121)}}{4}

k =\frac{-(-9) \pm 11}}{4}

k= \frac{ 9 \pm 11}{4}

k =  \frac{20}{4}                         k =  \frac{-2}{4}    

k_1 =5                                      k_2 = -\frac{1}{2}

2k^2-9k-5= 2(k-5)(k+\frac{1}{2})

Solving the RHS we get

\frac{2}{2}(k-5)(2k+1)

(k-5)(2k+1)

Substituting k = x+y

((x+y)-5)(2(x+y+1)

((x+y)-5)(2x+2y+1)

5 0
3 years ago
(1,3) (-1,0) and (5,0) (3,-3) parallel or perpendicular
nalin [4]
Parallel bc the slopes the same!
4 0
3 years ago
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