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Daniel [21]
3 years ago
12

Find the slope of the line that goes through the given points. (-2, -3) and ( - 14,11)

Mathematics
1 answer:
belka [17]3 years ago
7 0

Answer:

-7/8

Step-by-step explanation:

The formula for slope is [ y2-y1/x2-x1 ].

11-(-3)/-14-(-2)

14/-16

-7/8

Best of Luck!

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Simply the square root of negative 50
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What fractions are in between -1/3 any -2/3
UNO [17]

To find a fraction between two fractions, all we need to do is make the sum of the numerators be the new numerator, and the sum of the denominators be the new denominator.(x) So, for example, a fraction between 7/13 and 6/11 is (7 + 6)/(13+ 11) =13/24.(x)

7/13 = .5384615(x)

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13/24 = .541666�

 

Given that a/b < c/d, why is it true that a/b < (a+c)/(b+d)< c/d?

6 0
3 years ago
Spark plugs should be charged every 15,000 miles. if a truck runs 5,000 miles a month,how many times a year should the spark plu
Furkat [3]
One month = 5000

Number of months before the spark plugs need to change :

15000 ÷ 5000 = 3

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-----------------------------------------------------------------------------
Answer : It needs to be change 4 times a year.
-----------------------------------------------------------------------------
8 0
3 years ago
Х- а<br>x-b<br>If f(x) = b.x-a÷b-a + a.x-b÷a - b<br>Prove that: f (a) + f(b) = f (a + b)​
GenaCL600 [577]

Given:

Consider the given function:

f(x)=\dfrac{b\cdot(x-a)}{b-a}+\dfrac{a\cdot(x-b)}{a-b}

To prove:

f(a)+f(b)=f(a+b)

Solution:

We have,

f(x)=\dfrac{b\cdot(x-a)}{b-a}+\dfrac{a\cdot (x-b)}{a-b}

Substituting x=a, we get

f(a)=\dfrac{b\cdot(a-a)}{b-a}+\dfrac{a\cdot (a-b)}{a-b}

f(a)=\dfrac{b\cdot 0}{b-a}+\dfrac{a}{1}

f(a)=0+a

f(a)=a

Substituting x=b, we get

f(b)=\dfrac{b\cdot(b-a)}{b-a}+\dfrac{a\cdot (b-b)}{a-b}

f(b)=\dfrac{b}{1}+\dfrac{a\cdot 0}{a-b}

f(b)=b+0

f(b)=b

Substituting x=a+b, we get

f(a+b)=\dfrac{b\cdot(a+b-a)}{b-a}+\dfrac{a\cdot (a+b-b)}{a-b}

f(a+b)=\dfrac{b\cdot (b)}{b-a}+\dfrac{a\cdot (a)}{-(b-a)}

f(a+b)=\dfrac{b^2}{b-a}-\dfrac{a^2}{b-a}

f(a+b)=\dfrac{b^2-a^2}{b-a}

Using the algebraic formula, we get

f(a+b)=\dfrac{(b-a)(b+a)}{b-a}          [\because b^2-a^2=(b-a)(b+a)]

f(a+b)=b+a

f(a+b)=a+b               [Commutative property of addition]

Now,

LHS=f(a)+f(b)

LHS=a+b

LHS=f(a+b)

LHS=RHS

Hence proved.

5 0
2 years ago
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