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2 years ago

Pleaseeeeee show the work

Mathematics

1 answer:

trapecia [35]2 years ago

5 0

Answer:

I got x=10

Step-by-step explanation:

Step 1

Everything must add up to 180

180= 6x - 6 + 7x + 1 +55

Step 2

Combine everything

6x + 7x = 13x

-6 + 1 +55 =50

180 = 13x + 50

Step 4

Subtract. 180 -50

130 = 13x

Step 5

Divide 130/13

X=10

6(10) -6 = 54

7(10) + 1=71

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Solve -3/4(8k-24)>-12

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2 years ago

Find all points on the x-axis that are 14 units from the point (6,-7) All points on the x-axis that are 14 units from the point

Maksim231197 [3]

Answer: $(6+7\sqrt{3},0)\text{ and }(6-7\sqrt{3},0)$ are the required points.

or (18.124,0) and ( -6.124,0) are the required points.

Step-by-step explanation:

Let (x,0) be the point on x -axis that are 14 units from the point (6,-7) .

Then by distance formula , we have

$\sqrt{(x-6)^2+(0-(-7))^2}=14\ \ \ [\ \because distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}]$

Taking square on both the sides , we get

$(x-6)^2+7^2=14^2\\\\\Rightarrow\ x^2+6^2-2(6)x+49=196\\\\\Rightarrow\ x^2+36-12x=147\\\\\Rightarrow\ x^2-12x=111\\\\\Rightarrow\ x^2-12x-111=0$

Using quadratic formula : $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x=\dfrac{12\pm\sqrt{(-12)^2-4(1)(-111)}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm\sqrt{144+444}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm\sqrt{588}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm\sqrt{2^2\times7^2\times3}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm14\sqrt{3}}{2}\\\\\Rightarrow\ x=6\pm7\sqrt{3}$