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2 years ago

If m YXW=18°, YW=6, and WZ=6, what is ZXY?

Mathematics

1 answer:

MA_775_DIABLO [31]2 years ago

4 0

Answer:

18°

Step-by-step explanation:

It would also be 18 degrees

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Step-by-step explanation:

3x3

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2 years ago

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3 years ago

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2v^2-12 =-12v <br> Would really appreciate it loves❤️

Black_prince [1.1K]

For this case we have the following equation:

$2v ^ 2-12 = -12v$

Rewriting we have:

$2v ^ 2 + 12v-12 = 0$

Dividing by 2 to both sides of the equation:

$v ^ 2 + 6v-6 = 0$

We apply the quadratic formula:

$x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2a}$

We have to:

$a = 1\\b = 6\\c = -6$

Substituting:

$x = \frac {-6 \pm \sqrt {6 ^ 2-4 (1) (- 6)}} {2 (1)}\\x = \frac {-6 \pm \sqrt {36 + 24}} {2}\\x = \frac {-6 \pm \sqrt {60}} {2}\\x = \frac {-6 \pm \sqrt {4 * 15}} {2}\\x = \frac {-6 \pm2 \sqrt {15}} {2}$

Thus, we have two roots:

$x_ {1} = - 3+ \sqrt {15}\\x_ {2} = - 3- \sqrt {15}$

ANswer:

$x_ {1} = - 3+ \sqrt {15}\\x_ {2} = - 3- \sqrt {15}$

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3 years ago

Sharon wants to buy a shirt that costs $55. The sales tax is 5%. How much is the sales tax? What is her total cost for the shirt

Semmy [17]

easy

5% is .05 two decimals to the left

55 x .05 = 2.75

$2.75 is your sales tax

55+2.75=$57.75

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3 years ago

Approximate the change in the volume of a sphere when its radius changes from r = 40 ft to r equals 40.05 ft (Upper V (r )equal

alexgriva [62]

Answer:

The change in the volume of a sphere whose radius changes from 40 feet to 40.05 feet is approximately 1005.310 cubic feet.

Step-by-step explanation:

The volume of the sphere ( $V$ ), measured in cubic feet, is represented by the following formula:

$V = \frac{4\pi}{3}\cdot r^{3}$

Where $r$ is the radius of the sphere, measured in feet.

The change in volume is obtained by means of definition of total difference:

$\Delta V = \frac{\partial V}{\partial r}\Delta r$

The derivative of the volume as a function of radius is:

$\frac{\partial V}{\partial r} = 4\pi \cdot r^{2}$

Then, the change in volume is expanded:

$\Delta V = 4\pi \cdot r^{2}\cdot \Delta r$

If $r = 40\,ft$ and $\Delta r = 40\,ft-40.05\,ft = 0.05\,ft$ , the change in the volume of the sphere is approximately:

$\Delta V \approx 4\pi\cdot (40\,ft)^{2}\cdot (0.05\,ft)$

$\Delta V \approx 1005.310\,ft^{3}$

The change in the volume of a sphere whose radius changes from 40 feet to 40.05 feet is approximately 1005.310 cubic feet.

7 0

3 years ago