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BlackZzzverrR [31]
2 years ago
10

Minhaoh i need help go to my profile the qustion will be there

Mathematics
2 answers:
inessss [21]2 years ago
5 0

Answer:

1/4

Step-by-step explanation:

(-7,2) (1,4)

gtnhenbr [62]2 years ago
3 0

Answer:

What do you need help with

Step-by-step explanation:

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Write the rule to describe the translation.
lara31 [8.8K]

Answer:rqs

Step-by-step explanation:

7 0
3 years ago
Six equilateral triangles are connected to create a regular hexagon. The area of the hexagon is 24a^2 – 18 square units. Which i
netineya [11]

we know that

The area of the hexagon is equal to the sum of the areas of the six equilateral triangles

Let

x-------> area of one equilateral triangle

so

6x=24a^{2} -18

Divide by 6 both sides

x=4a^{2} -3 -------> area of one equilateral triangle

To find an equivalent expression for the area of the hexagon based on the area of a triangle, multiply the area of one equilateral triangle by 6

6*(4a^{2} -3)

therefore

the answer is

The equivalent expression is equal to 6*(4a^{2} -3)

6 0
3 years ago
Read 2 more answers
QUESTION 4
ANEK [815]

(2x-1)(x+4)=0

Step-by-step explanation:

A random zero property of multiplication is taken to find the solution

(2x-1)(x+4)=0

consider a=2x-1 and b=x-4                                      a.b=0

                                                                           either a or b or both must be 0

equating both the equations

2x-1=0 or x=4=0        x-4=0

2x-1=0                         x=4  

2x=1

x=1/2

substitute the values of x in the main equation

[2(1/2)-1][(1/2)+4]=0

3 0
3 years ago
Read 2 more answers
During one month there were seven days of precipitation. What if there had only been three days and precipitation that month? Ho
Mazyrski [523]
No se por que no te entendi
3 0
3 years ago
Please help!
erik [133]
For the answer to the question above, 
1 + nx + [n(n-1)/(2-factorial)](x)^2 + [n(n-1)(n-2)/3-factorial] (x)^3 

<span>1 + nx + [n(n-1)/(2 x 1)](x)^2 + [n(n-1)(n-2)/3 x 2 x 1] (x)^3 </span>

<span>1 + nx + [n(n-1)/2](x)^2 + [n(n-1)(n-2)/6] (x)^3 </span>

<span>1 + 9x + 36x^2 + 84x^3 </span>

<span>In my experience, up to the x^3 is often adequate to approximate a route. </span>

<span>(1+x) = 0.98 </span>

<span>x = 0.98 - 1 = -0.02 </span>

<span>Substituting: </span>

<span>1 + 9(-0.02) + 36(-0.02)^2 + 84(-0.02)^3 </span>

<span>approximation = 0.834 </span>

<span>Checking the real value in your calculator: </span>

<span>(0.98)^9 = 0.834 </span>

<span>So you have approximated correctly. </span>

<span>If you want to know how accurate your approximation is, write out the result of each in full: </span>

<span>1 + 9(-0.02) + 36(-0.02)^2 + 84(-0.02)^3 = 0.833728 </span>

<span> (0.98)^9 = 0.8337477621 </span>

<span>So it is correct to 4</span>
5 0
2 years ago
Read 2 more answers
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