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2 years ago

Find the value of X to the nearest tenth:

Mathematics

1 answer:

castortr0y [4]2 years ago

5 0

Answer:

13.0

Step-by-step explanation:

sin(65°) × 14.3 would give u 12.9 then u round it to the 13

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15 points!

AveGali [126]

Answer:

Since we are assuming a constant velocity, we know that v = d/t which means we can rearrange this formula to solve for distance:

d = v * t

d = (20 cm/sec) * (25 sec) = 500 cm

Step-by-step explanation:

<h2><u><em>BRAINLIEST?</em></u></h2>

6 0

2 years ago

Please help me, look at the picture I need help... please

Umnica [9.8K]

Answer: Funciton A is linear. Its equation is y=1.66~(x) + 0

Step-by-step explanation:

See if this works. I’m not 100% sure!

7 0

2 years ago

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Express 24 out of 80 as a percent

olga2289 [7]

Answer:

30%

Step-by-step explanation:

24/80 = 0.3

move the decimal over 2 times and you'll get your answer

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3 years ago

97÷2,134 and how to work it out

soldi70 [24.7K]

Answer:

The answer is either 9/200 (nine two-hundreths) or 22

3 0

3 years ago

Read 2 more answers

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

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

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

since $\sqrt{3}=1.732$

so, $(6+7(1.732),0)\text{ and }(6-7(1.732),0)$ are the required points.

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

3 0

3 years ago