Ask question

Ask question

All categories

anyanavicka [17]

3 years ago

13

Ellen walks dogs on weekends. She gets paid $8.50 per hour. Each day she works five hours and 45 minutes. How much does Ellen ge

t paid per weekend?

1 answer:

Nikitich [7]3 years ago

5 0

48.875 i believe is the answer hope it helps!!!

You might be interested in

What can be the next step please helppppppppppppp

stepladder [879]

Answer:

Statement: Triangle ACD is congruent to Triangle BCD

Reason: SSA (Side, Side, Angle)

8 0

3 years ago

<img src="https://tex.z-dn.net/?f=5%20%5Cdiv%202%20%5Csqrt%7B2%7D%20" id="TexFormula1" title="5 \div 2 \sqrt{2} " alt="5 \div 2

Rufina [12.5K]

$\dfrac{5}{2 \sqrt{2} }$

it's the only ans then possible ! ^•,,,•^

7 0

2 years ago

Read 2 more answers

Help me please! I'm so terrible at math

Xelga [282]

You have to take log 10/ log 3 and round to the two decimal places. Your answer should be c- 0.48

5 0

3 years ago

Read 2 more answers

For all integers n, if n2 is odd, then n is odd. Use a proof by contraposition, as in Lemma 1.1.

Mumz [18]

<span>contraP is to flip and negate the prepositions right? if A then B, if -B then -A

Can I have Brainlest please?

</span>

5 0

3 years ago

Find the equation of the sphere if one of its diameters has endpoints (4, 2, -9) and (6, 6, -3) which has been normalized so tha

Pavel [41]

Answer:

$(x - 5)^2 + (y - 4)^2 + (z - 6)^2 = 14$ .

(Expand to obtain an equivalent expression for the sphere: $x^2 - 10\,x + y^2 - 8\, y + z^2 - 12\, z + 63 = 0$ )

Step-by-step explanation:

Apply the Pythagorean Theorem to find the distance between these two endpoints:

$\begin{aligned}&\text{Distance}\cr &= \sqrt{\left(x_2 - x_1\right)^2 + \left(y_2 - y_1\right)^2 + \left(z_2 - z_1\right)^2} \cr &= \sqrt{(6 - 4)^2 + (6 - 2)^2 + ((-3) - (-9))^2 \cr &= \sqrt{56}}\end{aligned}$ .

Since the two endpoints form a diameter of the sphere, the distance between them would be equal to the diameter of the sphere. The radius of a sphere is one-half of its diameter. In this case, that would be equal to:

$\begin{aligned} r &= \frac{1}{2} \, \sqrt{56} \cr &= \sqrt{\left(\frac{1}{2}\right)^2 \times 56} \cr &= \sqrt{\frac{1}{4} \times 56} \cr &= \sqrt{14} \end{aligned}$ .

In a sphere, the midpoint of every diameter would be the center of the sphere. Each component of the midpoint of a segment (such as the diameter in this question) is equal to the arithmetic mean of that component of the two endpoints. In other words, the midpoint of a segment between $\left(x_1, \, y_1, \, z_1\right)$ and $\left(x_2, \, y_2, \, z_2\right)$ would be:

$\displaystyle \left(\frac{x_1 + x_2}{2},\, \frac{y_1 + y_2}{2}, \, \frac{z_1 + z_2}{2}\right)$ .

In this case, the midpoint of the diameter, which is the same as the center of the sphere, would be at:

$\begin{aligned}&\left(\frac{x_1 + x_2}{2},\, \frac{y_1 + y_2}{2}, \, \frac{z_1 + z_2}{2}\right) \cr &= \left(\frac{4 + 6}{2},\, \frac{2 + 6}{2}, \, \frac{(-9) + (-3)}{2}\right) \cr &= (5,\, 4\, -6)\end{aligned}$ .

The equation for a sphere of radius $r$ and center $\left(x_0,\, y_0,\, z_0\right)$ would be:

$\left(x - x_0\right)^2 + \left(y - y_0\right)^2 + \left(z - z_0\right)^2 = r^2$ .

In this case, the equation would be:

$\left(x - 5\right)^2 + \left(y - 4\right)^2 + \left(z - (-6)\right)^2 = \left(\sqrt{56}\right)^2$ .

Simplify to obtain:

$\left(x - 5\right)^2 + \left(y - 4\right)^2 + \left(z + 6\right)^2 = 56$ .

Expand the squares and simplify to obtain:

$x^2 - 10\,x + y^2 - 8\, y + z^2 - 12\, z + 63 = 0$ .

8 0

3 years ago