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3 years ago

10

A baker uses a coffee mug with a diameter of 8cm to cut out circular cookies from a big sheet of cookie dough. What is the area

A of each cookie? Give your answer in terms of pi.

1 answer:

kodGreya [7K]3 years ago

4 0

Diameter = 8 cm

radius = diameter/2 = 4 cm

A = pi * r^2

A = pi * (4 cm)^2

A = 16pi cm^2

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Given points A (1, 2/3), B (x, -4/5), and C (-1/2, 4) determine the value of x such that all three points are collinear

AlladinOne [14]

Answer:

$x=\frac{83}{50}$

Step-by-step explanation:

we know that

If the three points are collinear

then

$m_A_B=m_A_C$

we have

A (1, 2/3), B (x, -4/5), and C (-1/2, 4)

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

step 1

Find the slope AB

we have

$A(1,\frac{2}{3}),B(x,-\frac{4}{5})$

substitute in the formula

$m_A_B=\frac{-\frac{4}{5}-\frac{2}{3}}{x-1}$

$m_A_B=\frac{\frac{-12-10}{15}}{x-1}$

$m_A_B=-\frac{22}{15(x-1)}$

step 2

Find the slope AC

we have

$A(1,\frac{2}{3}),C(-\frac{1}{2},4)$

substitute in the formula

$m_A_C=\frac{4-\frac{2}{3}}{-\frac{1}{2}-1}$

$m_A_C=\frac{\frac{10}{3}}{-\frac{3}{2}}$

$m_A_C=-\frac{20}{9}$

step 3

Equate the slopes

$m_A_B=m_A_C$

$-\frac{22}{15(x-1)}=-\frac{20}{9}$

solve for x

$15(x-1)20=22(9)$

$300x-300=198$

$300x=198+300$

$300x=498$

$x=\frac{498}{300}$

simplify

$x=\frac{83}{50}$

8 0

4 years ago

Which expression is equivalent to cot t sec t?

Solnce55 [7]

Answer: $csc(t)$

Step-by-step explanation:

Alright, lets get started.

The given expression is given as :

$cot( t) \times sec (t)$

We know quotient identity as :

$cot(t)=\frac{cos(t)}{sin(t)}$

Similarly, we know reciprocal identity as :

$sec(t)=\frac{1}{cos(t)}$

lets plug the value of cot and sec in given expression

$\frac{cos(t)}{sin(t)} \times \frac{1}{cos(t)}$

cos will be cancelled, remaining will be

$\frac{1}{sin(t)}$

Using reciprocal identity again, that will equal to :

$csc(t)$ ................... Answer (A)

8 0

3 years ago

Daniel earns $760 each month he pays 15% of his earning in tax.

Gwar [14]

15 percent of 760 is 114
760 - 114 = 646

Hope this helps :D

6 0

3 years ago

The diagram shows a semi circle inside a rectangle of length 150 m. The semi circle touches the rectangle at a b and c. Calculat

belka [17]

Answer:

Step-by-step explanation:

<h3>C=π×d=3.142×160=502.72 </h3><h3>502.72/4=125.68cm </h3><h3>Because we can make from the rectangle two squares and then we would have 4 squares in total.</h3><h3>That is for 2 marks.If you have got any ideas for the other two marks,I would love to hear from you.I am doing the same question.Also,I see you have done question 7.Would you mind to help me with it?</h3>

5 0

3 years ago

Read 2 more answers

If X and Y are independent continuous positive random

Leni [432]

a) $Z=\frac XY$ has CDF

$F_Z(z)=P(Z\le z)=P(X\le Yz)=\displaystyle\int_{\mathrm{supp}(Y)}P(X\le yz\mid Y=y)P(Y=y)\,\mathrm dy$

$F_Z(z)\displaystyle=\int_{\mathrm{supp}(Y)}P(X\le yz)P(Y=y)\,\mathrm dy$

where the last equality follows from independence of $X,Y$ . In terms of the distribution and density functions of $X,Y$ , this is

$F_Z(z)=\displaystyle\int_{\mathrm{supp}(Y)}F_X(yz)f_Y(y)\,\mathrm dy$

Then the density is obtained by differentiating with respect to $z$ ,

$f_Z(z)=\displaystyle\frac{\mathrm d}{\mathrm dz}\int_{\mathrm{supp}(Y)}F_X(yz)f_Y(y)\,\mathrm dy=\int_{\mathrm{supp}(Y)}yf_X(yz)f_Y(y)\,\mathrm dy$

b) $Z=XY$ can be computed in the same way; it has CDF

$F_Z(z)=P\left(X\le\dfrac zY\right)=\displaystyle\int_{\mathrm{supp}(Y)}P\left(X\le\frac zy\right)P(Y=y)\,\mathrm dy$

$F_Z(z)\displaystyle=\int_{\mathrm{supp}(Y)}F_X\left(\frac zy\right)f_Y(y)\,\mathrm dy$

Differentiating gives the associated PDF,

$f_Z(z)=\displaystyle\int_{\mathrm{supp}(Y)}\frac1yf_X\left(\frac zy\right)f_Y(y)\,\mathrm dy$

Assuming $X\sim\mathrm{Exp}(\lambda_x)$ and $Y\sim\mathrm{Exp}(\lambda_y)$ , we have

$f_{Z=\frac XY}(z)=\displaystyle\int_0^\infty y(\lambda_xe^{-\lambda_xyz})(\lambda_ye^{\lambda_yz})\,\mathrm dy$

$\implies f_{Z=\frac XY}(z)=\begin{cases}\frac{\lambda_x\lambda_y}{(\lambda_xz+\lambda_y)^2}&\text{for }z\ge0\\0&\text{otherwise}\end{cases}$

and

$f_{Z=XY}(z)=\displaystyle\int_0^\infty\frac1y(\lambda_xe^{-\lambda_xyz})(\lambda_ye^{\lambda_yz})\,\mathrm dy$

$\implies f_{Z=XY}(z)=\lambda_x\lambda_y\displaystyle\int_0^\infty\frac{e^{-\lambda_x\frac zy-\lambda_yy}}y\,\mathrm dy$

I wouldn't worry about evaluating this integral any further unless you know about the Bessel functions.

6 0

3 years ago