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3 years ago

The function q(w)=3+5(w−1) represents the number of quarters in a bowl on week w.

Mathematics

2 answers:

olga2289 [7]3 years ago

5 0

Answer:

Five quarters are added to the bowl every week.

mamaluj [8]3 years ago

4 0

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
The value 5 represent in this situation is letter B which is "
<span>The value of the quarters in the bowl on Week 1 was $5."</span>

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Kurt needs pounds of chocolate chips to make cookies. How many ounces of chocolate chips does he need? PLEASE I NEED IT RIGHT AW

Bad White [126]

Answer:

Step-by-step explanation:

Multiply the number of pounds by 16.

1 LB=16OZ

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Triangle BCD is isosceles and line Bc is congruent to line BD. What is the measure of arc BC?

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Since arc CD is 100 degrees, the rest of the circle must be 260 degrees. Since the rest of the circle is made up of two equal arcs, then arc BC must be half of 260 degrees. Thus, the answer is 130 degrees, C.

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3 years ago

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Kite EFGH is inscribed in a rectangle where F and H are midpoints of parallel sides. The area of EFGH is 35 square units. What i

sasho [114]

*see attachment for the figure described

Answer:

5 units

Step-by-step explanation:

==>Given the figure attached below, let where FH and EG intercepted be K.

Since FH are midpoints of parallel lines, KE = KG = x.

Given that the area of the kite EFGH = 35 square units, and we know the length of one of the diagonals = HF = KF + KH = 2 + 5 = 7, we can solve for x using the formula for the area of a kite.

Area of kite = ½ × d1 × d2

Where d1 = KH = 7

d2 = EG = KE + KG = x + x = 2x

Area of kite EFGH = 35

THUS:

35 = ½ × 7 × 2x

35 = 1 × 7 × x

35 = 7x

Divide both sides by 7

35/7 = x

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4 0

3 years ago

Find an equation of the tangent plane to the given parametric surface at the specified point.

Neko [114]

Answer:

Equation of tangent plane to given parametric equation is:

$\frac{\sqrt{3}}{2}x-\frac{1}{2}y+z=\frac{\pi}{3}$

Step-by-step explanation:

Given equation

$r(u, v)=u cos (v)\hat{i}+u sin (v)\hat{j}+v\hat{k}$ ---(1)

Normal vector tangent to plane is:

$\hat{n} = \hat{r_{u}} \times \hat{r_{v}}\\r_{u}=\frac{\partial r}{\partial u}\\r_{v}=\frac{\partial r}{\partial v}$

$\frac{\partial r}{\partial u} =cos(v)\hat{i}+sin(v)\hat{j}\\\frac{\partial r}{\partial v}=-usin(v)\hat{i}+u cos(v)\hat{j}+\hat{k}$

Normal vector tangent to plane is given by:

$r_{u} \times r_{v} =det\left[\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\cos(v)&sin(v)&0\\-usin(v)&ucos(v)&1\end{array}\right]$

Expanding with first row

$\hat{n} = \hat{i} \begin{vmatrix} sin(v)&0\\ucos(v) &1\end{vmatrix}- \hat{j} \begin{vmatrix} cos(v)&0\\-usin(v) &1\end{vmatrix}+\hat{k} \begin{vmatrix} cos(v)&sin(v)\\-usin(v) &ucos(v)\end{vmatrix}\\\hat{n}=sin(v)\hat{i}-cos(v)\hat{j}+u(cos^{2}v+sin^{2}v)\hat{k}\\\hat{n}=sin(v)\hat{i}-cos(v)\hat{j}+u\hat{k}\\$

at u=5, v =π/3

$=\frac{\sqrt{3} }{2}\hat{i}-\frac{1}{2}\hat{j}+\hat{k}$ ---(2)

at u=5, v =π/3 (1) becomes,

$r(5, \frac{\pi}{3})=5 cos (\frac{\pi}{3})\hat{i}+5sin (\frac{\pi}{3})\hat{j}+\frac{\pi}{3}\hat{k}$

$r(5, \frac{\pi}{3})=5(\frac{1}{2})\hat{i}+5 (\frac{\sqrt{3}}{2})\hat{j}+\frac{\pi}{3}\hat{k}$

$r(5, \frac{\pi}{3})=\frac{5}{2}\hat{i}+(\frac{5\sqrt{3}}{2})\hat{j}+\frac{\pi}{3}\hat{k}$

From above eq coordinates of r₀ can be found as:

$r_{o}=(\frac{5}{2},\frac{5\sqrt{3}}{2},\frac{\pi}{3})$

From (2) coordinates of normal vector can be found as

$n=(\frac{\sqrt{3} }{2},-\frac{1}{2},1)$

Equation of tangent line can be found as:

$(\hat{r}-\hat{r_{o}}).\hat{n}=0\\((x-\frac{5}{2})\hat{i}+(y-\frac{5\sqrt{3}}{2})\hat{j}+(z-\frac{\pi}{3})\hat{k})(\frac{\sqrt{3} }{2}\hat{i}-\frac{1}{2}\hat{j}+\hat{k})=0\\\frac{\sqrt{3}}{2}x-\frac{5\sqrt{3}}{4}-\frac{1}{2}y+\frac{5\sqrt{3}}{4}+z-\frac{\pi}{3}=0\\\frac{\sqrt{3}}{2}x-\frac{1}{2}y+z=\frac{\pi}{3}$

5 0

2 years ago