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uranmaximum [27]
3 years ago
7

Solve the proportion for k k+7 = 4k+15 29= 21

Mathematics
2 answers:
OleMash [197]3 years ago
5 0

Answer:

k=3/8

Step-by-step explanation:

k+7 = 4k+15

-k       -k

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

7=3k +15

-15    -15

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

8=3k

----------

3

k=3/8

mafiozo [28]3 years ago
4 0

Answer:

k+7

k-7

or k+7/k-7

Step-by-step explanation:

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Which equation represents a line which is parallel to the line 2x+y=-82x+y=−8?
geniusboy [140]

Answer:

x=7 is the line parallel to the given line x=8 . Parallel lines are the lines which are always same distance apart and never intersect. The given line is x=8 with intersects the x axis at x=8 and it is parallel to y axis with slope infinite or not defined.

Step-by-step explanation:

6 0
2 years ago
What is it pls help !
Nesterboy [21]

Answer:

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6 0
2 years ago
How long does it take for a simple with total number n to decay to 1/3n?
Vlada [557]
<span>Income earned over one year is principal * int rate so we have 0.12*A + 0.13*B +0.14*C = 400 and A + B + C = 3000 and A+B = C</span>
7 0
3 years ago
Ben weighs 140.8 pounds, and Jennifer weighs 117.7 pounds. How many more pounds does Ben weigh than Jennifer?
DENIUS [597]

Answer:

Ben weighs 23.1 lbs more than Jennifer.

Step-by-step explanation:

To find the difference between Ben and Jennifer's weight, you have to subtract the smaller number from the bigger number. In this case, subtract Jennifer's weight from Ben's weight. This would look like this: 140.8-117.8.

140.8-117.7=23.1

Therefore, Ben weighs 23.1 lbs more than Jennifer.

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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