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3 years ago

11

Pleasseeeee helpppppppppppppp hurryyy please

1 answer:

Arlecino [84]3 years ago

4 0

So to answer all of them ( I don’t know if you want that it Imma do it anyways). For the first question: Name a radius - QR. For the second question: Name a diameter - NR. For question three: Name a chord - OP. For the last question: If the length of QS is 8 units, what is the length of NR: Since QS is a radius than the length of NR would be 16. Hope this helps!

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Quadrilateral ABCD is rotated 90° clockwise about the origin. What are the

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A rectangular piece of land measures 20 feet by 15 feet. James created a cement sidewalk, x feet wide, to border a rectangular g

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Step-by-step explanation:

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Solve for u<br> 6≤ -9/4u

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$6 \leqslant \frac{4u}{ - 9}$

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A cylindrical hole is bored through a steel block, and a cylindrical piston is machined to fit into the hole. The diameter of th

prisoha [69]

Answer:

The diameter of the hole is 20.04 ± 0.01 cm

The diameter of the piston is 19.92 ± 0.04 cm.

We are given that The clearance is one-half the difference between the diameters.

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Uncertainty in the estimate = $\frac{1}{2}(0.01+0.04)$

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3 0

3 years ago

Find the general solution to 1/x dy/dx - 2y/x^2 = x cos x, y(pi) = pi^2

Finger [1]

Answer:

$\frac{y}{x^2}=\sin x+\pi$

Step-by-step explanation:

Consider linear differential equation $\frac{\mathrm{d} y}{\mathrm{d} x}+yp(x)=q(x)$

It's solution is of form $y\,I.F=\int I.F\,q(x)\,dx$ where I.F is integrating factor given by $I.F=e^{\int p(x)\,dx}$ .

Given: $\frac{1}{x}\frac{\mathrm{d} y}{\mathrm{d} x}-\frac{2y}{x^2}=x\cos x$

We can write this equation as $\frac{\mathrm{d} y}{\mathrm{d} x}-\frac{2y}{x}=x^2\cos x$

On comparing this equation with $\frac{\mathrm{d} y}{\mathrm{d} x}+yp(x)=q(x)$ , we get $p(x)=\frac{-2}{x}\,\,,\,\,q(x)=x^2\cos x$

I.F = $e^{\int p(x)\,dx}=e^{\int \frac{-2}{x}\,dx}=e^{-2\ln x}=e^{\ln x^{-2}}=\frac{1}{x^2}$ { formula used: $\ln a^b=b\ln a$ }

we get solution as follows:

$\frac{y}{x^2}=\int \frac{1}{x^2}x^2\cos x\,dx\\\frac{y}{x^2}=\int \cos x\,dx\\\\\frac{y}{x^2}=\sin x+C$

{ formula used: $\int \cos x\,dx=\sin x$ }

Applying condition: $y(\pi)=\pi^2$

$\frac{y}{x^2}=\sin x+C\\\frac{\pi^2}{\pi}=\sin\pi+C\\\pi=C$

So, we get solution as :

$\frac{y}{x^2}=\sin x+\pi$

4 0

3 years ago