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3 years ago

Pls guys help me pls :c

Mathematics

1 answer:

Vinil7 [7]3 years ago

8 0

Answer: A. $x\geq21\dfrac{2}{3}$ .

Step-by-step explanation:

The given inequation: $81-1\dfrac{1}{5}x\leq55$

$\Rightarrow\ 81-\dfrac{6}{5}x\leq55$

Multiplying 5 on both the sides, we get

$5(81-\dfrac{6}{5}x)\leq5\times55\\\\\Rightarrow\ 405-6x\leq275\\\\\Rightarrow\ 6x\geq405-275\\\\\Rightarrow\ 6x\geq130\\\\\Rightarrow\ x\geq\dfrac{130}{6}\\\\\Rightarrow\ x\geq\dfrac{65}{3}\\\\\Rightarrow\ x\geq21\dfrac{2}{3}$

Hence, the inequality for x is $x\geq21\dfrac{2}{3}$ .

So, the correct option is A. $x\geq21\dfrac{2}{3}$ ..

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Step-by-step explanation:

We want to find the point in QI at which the line with the equation:

$y=0.5x+5$

Intersect a circle with a radius of 3 and a center of (0, 5).

First, write the equation of a circle. The equation for a circle is given by:

$(x-h)^2+(y-k)^2=r^2$

Where (h, k) is the center and r is the radius.

Since our center is (0, 5), h = 0 and k = 5. The radius is 3. So, r = 3. Substitute:

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$x^2+(y-5)^2=9$

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$x^2+((0.5x+5)-5)^2=9$

Simplify:

$x^2+(0.5x)^2=9$

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$x^2+0.25x^2=9$

Combine like terms:

$\displaystyle 1.25x^2=\frac{5}{4}x^2=9$

Solve for x:

$\displaystyle \begin{aligned} x^2&=\frac{36}{5} \\ x&=\pm\sqrt{\frac{36}{5}} \\ x&\Rightarrow \frac{6}{\sqrt{5}}=\frac{6\sqrt{5}}{5}\approx2.68\end{aligned}$

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To find the y-coordinate, substitute the x-value back into either equation. Using the linear equation:

$\displaystyle y=0.5\left(\frac{6\sqrt{5}}{5}\right)+5=\frac{3\sqrt{5}}{5}+5\approx 6.34$

So, the point of intersection in QI is:

$\displaystyle \left(\frac{6\sqrt{5}}{5}, \frac{3\sqrt{5}}{5}+5\right)\approx \left(2.68, 6.34)$

6 0

3 years ago

Pls guys help me pls :c​

Pls guys help me pls :c