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3 years ago

8 1/5 - 2 1/3 I need to find what this answer is quick! please help me!

Mathematics

2 answers:

velikii [3]3 years ago

7 0

The answer is 5 13/15

kondaur [170]3 years ago

5 0

The answer is 5 13/15 or atleast according to calculator.net (the website used in the picture) :)

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Select the mathematical expression that corresponds to the following written

mestny [16]

Answer:

none of the above

Step-by-step explanation:

(t÷5) - 25

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3 years ago

PLEASE HELP ME WITH THIS MATH QUESTION

Triss [41]

Answer:

Permutations and combinations

Step-by-step explanation:

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3 years ago

Which line has a slope of 12 and a y-intercept of 2?

tino4ka555 [31]

Answer:

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Step-by-step explanation:

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3 years ago

HELP ITS REALLY URGENT

SSSSS [86.1K]

Answer/Step-by-step explanation:

a. m<h = x°

Two angles to use in finding x are <h and 55°

Angle pair formed: linear pair angles

b. x° + 55° = 180° (linear pair)

Subtract 55° from each side

x + 55 - 55 = 180 - 55

x = 125°

c. Given: m<n = y°

Two angles to use in solving for y are: <h and <n

Angle pair formed: alternate exterior angles

d. m<n = y° (given)

m<n = x° = 125° (as solved in 2b)

Therefore:

y = x (alternate exterior angles are congruent)

thus:

y = 125° (substitution)

5 0

3 years ago

Anybody help me to solve this question.

Mumz [18]

Answer:

$\dfrac{1}{(b-c)},\dfrac{1}{(c-a)} ,\dfrac{1}{(a-b)} are\ in\ AP$

Step-by-step explanation:

Given that $(b-c)^2, (c-a)^2 , (a-b)^2$ are in AP

To prove: $\dfrac{1}{(b-c)},\dfrac{1}{(c-a)} ,\dfrac{1}{(a-b)}$ are in AP

From given as we know if p , q, r are in AP then 2q= p+r.

$2(c-a)^2= (b-c)^2+(a-b)^2\\\\\Rightarrow 2(c^2+a^2-2ac)=b^2+c^2-2bc+a^2+b^2-2ab\\\\\Rightarrow 2c^2+2a^2-4ac= 2b^2+c^2+a^2 -2bc-2ab\\\\\Rightarrow a^2+c^2-2b^2-4ac= -2bc-2ab\\\\\Rightarrow a^2-2b^2+c^2= 4ac-2bc-2ab$

Now

$\dfrac{1}{(b-c)},\dfrac{1}{(c-a)} ,\dfrac{1}{(a-b)}2\dfrac{1}{(c-a)} =\dfrac{1}{(b-c)}+\dfrac{1}{(a-b)}\\\\\Rightarrow \dfrac{2}{(c-a)}= \dfrac{a-b+b-c}{(b-c)(a-b)} \\\\\Rightarrow \dfrac{2}{(c-a)}= \dfrac{a-c}{(b-c)(a-b)} \\\\\Rightarrow2(b-c)(a-b) = (c-a)(a-c) \\\\\Rightarrow 2(ab-b^2-ac+bc)= -(a-c)^2\\\\\Rightarrow 2ab- 2b^2-2ac+2bc = -a^2-c^2+2ac\\\\\Rightarrow a^2-2b^2+c^2=4ac-2ab-2bc$

Which is the result of AP

Hence proved

6 0

3 years ago