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Lady_Fox
23 hours ago
5

the shorter sides of an acute triangle are x cm and 2x cm. the longest side of the trinalge is 15 cm. what is the smallest possi

ble whole-number value of x ?
Mathematics
1 answer:
Leona [9.2K]23 hours ago
3 0
The smallest whole-number value for x that works is 7. The triangle’s sides can be defined as: a = x, b = 2x, and c = 15. Recognizing c as the longest side leads us to the condition for an acute triangle: c^2 must be less than a^2 + b^2. Inputting the known values, we solve for x and find that x must exceed 6.708. As a result, the least integer that satisfies this requirement is 7.
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The salt enters at a rate of (5 g/L)*(3 L/min) = 15 g/min.

The salt exits at a rate of (x/10 g/L)*(3 L/min) = 3x/10 g/min.

Thus, the total rate of salt flow, represented by x(t) in grams, is defined by the differential equation,

x'(t)=15-\dfrac{3x(t)}{10}

which is linear. Shift the x term to the right side, then multiply both sides by e^{3t/10}:

e^{3t/10}x'+\dfrac{3e^{t/10}}{10}x=15e^{3t/10}

\implies\left(e^{3t/10}x\right)'=15e^{3t/10}

Next, integrate both sides and solve for x:

e^{3t/10}x=50e^{3t/10}+C

\implies x(t)=50+Ce^{-3t/10}

Initially, the tank contains 5 g of salt at time t=0, so we have

5=50+C\implies C=-45

\implies\boxed{x(t)=50-45e^{-3t/10}}

The duration required for the tank to contain 20 g of salt is t, such that

20=50-45e^{-3t/10}\implies t=\dfrac{20}3\ln\dfrac32\approx2.7031\,\mathrm{min}

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A rectangle is transformed according to the rule r0, 90º. the image of the rectangle has vertices located at r'(–4, 4), s'(–4, 1
Svet_ta [9500]

Answer:

For a 90º clockwise rotation, the vertex q' (–3, 4) maps to q(4, 3).

According to the 90º counterclockwise rotation rule, q' (–3, 4) transforms to q(–4, –3).

Step-by-step explanation:

Given the rectangle with vertices r'(–4, 4), s'(–4, 1), p'(–3, 1), and q'(–3, 4),

Find the image of vertex q after a 90º rotation.

The rotation rules are:

90º clockwise: (x, y) → (y, –x)

90º counterclockwise: (x, y) → (–y, x)

Applying the clockwise rule to q'(–3, 4) yields q(4, 3).

Applying the counterclockwise rule to q'(–3, 4) yields q(–4, –3).

Thus, clockwise rotation of q' (–3, 4) results in q(4, 3),

and counterclockwise rotation results in q(–4, –3).

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