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3 years ago

Please show process!!!! THANK YOU! Will mark brainylist

Mathematics

1 answer:

lakkis [162]3 years ago

3 0

Answer: 103 degrees

Step-by-step explanation:

51 + 26 = 77

A triangle adds up to 180 degrees

180 - 77 = 103

= 103 degrees

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I think the answer is C. 45 units

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3 years ago

Giving brainliest but has to be a actual answer

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2,800 Books to be the same.

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3 years ago

What is the equation to the line that goes through the point

DochEvi [55]

Step-by-step explanation:

perpendicular slope is negative reciprocal. -1/-1 = 1

y - 4 = x +3

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3 years ago

Edward is collecting data on the average length of lizards from head to tail for a particular generation. Initially, the average

Zina [86]

Answer:

The general form of a natural logarithmic function is f(x)=a In(x-h) +k. Since the initial average length of the lizards was 20 centimeters, one possible point (h+1,k) through which the function passes is (0,20).

In this case, h= -1 since h +1 = 0 and k = 20.

It follows that x - h = (-1) = x + 1.

So, the function takes the form f(x) = a In(x + 1) +20.

The average length of the lizards after 2 years is 22.197. So f(2) =22.197. We substitute this value into the function

Step-by-step explanation:

f(2) = a In(2 +1) +20

22.197 = a In(3) + 20

$\frac{2.197}{1.099}$ ≈ a

2 ≈ a

So the function that represents the average length of the lizards across this generation is f(x) = 2 In(x+1) +20

3 0

3 years ago

What is the solution for the equation StartFraction 5 Over 3 b cubed minus 2 b squared minus 5 EndFraction = StartFraction 2 Ove

wolverine [178]

Answer:

The solutions are:

$b=0,\:b=4$

Step-by-step explanation:

Considering the expression

$\frac{5}{3b^3-2b^2-5}=\frac{2}{b^3-2}$

Solving the expression

$\frac{5}{3b^3-2b^2-5}=\frac{2}{b^3-2}$

$\mathrm{Apply\:fraction\:cross\:multiply:\:if\:}\frac{a}{b}=\frac{c}{d}\mathrm{\:then\:}a\cdot \:d=b\cdot \:c$

$5\left(b^3-2\right)=\left(3b^3-2b^2-5\right)\cdot \:2$

$5b^3-10=6b^3-4b^2-10$

$\mathrm{Switch\:sides}$

$6b^3-4b^2-10=5b^3-10$

$6b^3-4b^2-10+10=5b^3-10+10$

$6b^3-4b^2=5b^3$

$\mathrm{Subtract\:}5b^3\mathrm{\:from\:both\:sides}$

$6b^3-4b^2-5b^3=5b^3-5b^3$

$b^3-4b^2=0$

$Using\:the\:Zero\:Factor\:Principle:$ $if\:\mathrm ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)$

So,

$b=0,b-4=0$

$b=0,b=4$

Therefore, the solutions are:

$b=0,\:b=4$

4 0

3 years ago

Read 2 more answers