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3 years ago

Shape A and shape B are each made from five identical squares.

Mathematics

2 answers:

kondaur [170]3 years ago

8 0

Answer:

Step-by-step explanation:

Answer= 50

mylen [45]3 years ago

6 0

Answer:

the answer is 60

Step-by-step explanation:

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What is the area of this figure?

Verdich [7]

Answer:

In the given composite figure:

Labelled the figure:

In a square ABCD figure:

Each side is of 18 m each

Area of a square is given by:

$\text{Area of square} = (\text{Side})^2$

Then;

$\text{Area of square ABCD} = (18)^2 = 324 m^2$

Now, In right angle triangle CEF:

Area of a right angle triangle is given by:

$\text{Area of triangle} = \frac{1}{2}(\text{Base}) \cdot (\text{Height})$

In a a triangle CEF:

Base = CE = 8 m and Height = CF = 16 m

then;

$\text{Area of triangle CEF} = \frac{1}{2}(8) \cdot (16)=4 \cdot 16 = 64 m^2$

Similarly in triangle DCF:-

Base = CD = 18 m

Height = CF = 16 m

then;

$\text{Area of triangle DCF} = \frac{1}{2}(18) \cdot (16)=9 \cdot 16 = 144 m^2$

Thus:

area of composite figure =Area of square ABCD + Area of triangle CEF + Area of triangle DCF

Area of this composite figure= 324 +64+144 = 532 square meter.

Therefore, area of this figure is 532 meter square.

7 0

3 years ago

The temperature at a point (x, y, z) is given by t(x, y, z) = 200e−x2 − 5y2 − 9z2 where t is measured in °c and x, y, z in meter

Olin [163]

Part A:

Given that

$T(x, y, z) = 200e^{-x^2-5y^2-9z^2} \\ \\ \nabla T=\left( \frac{\partial}{\partial x} +\frac{\partial}{\partial y}+\frac{\partial}{\partial z}\right)T=200e^{-x^2-5y^2-9z^2}\left\ \textless \ -2x,-10y,-18z\right\ \textgreater \ \\ \\ \nabla T(4,-1,4)=200e^{-(4)^2-5(-1)^2-9(4)^2}\left\ \textless \ -2(4),-10(-1),-18(4)\right\ \textgreater \ \\ \\ =200e^{-165}\left\ \textless \ -8,10,-72\right\ \textgreater \$

For the direction, let u be unit vector of point (6, -3, 6), then

$u= \frac{1}{||\left\ \textless \ 6,-3,6\right\ \textgreater \ ||} \left\ \textless \ 6,-3,6\right\ \textgreater \ \\ \\ = \frac{1}{\sqrt{6^2+(-3)^2+6^2}} \left\ \textless \ 6,-3,6\right\ \textgreater \ \\ \\ = \frac{1}{\sqrt{81}} \left\ \textless \ 6,-3,6\right\ \textgreater \ = \frac{1}{9} \left\ \textless \ 6,-3,6\right\ \textgreater \ \\ \\ =\frac{1}{3} \left\ \textless \ 2,-1,2\right\ \textgreater \$

Thus, the directional derivative is given by:

$D_u(4,-1,4)=200e^{-165}\left\ \textless \ -8,10,-72\right\ \textgreater \ \cdot\frac{1}{3} \left\ \textless \ 2,-1,2\right\ \textgreater \ \\ \\ = \frac{200}{3} e^{-165}(-16-10-144)=-\frac{170(200)}{3} e^{-165}\approx-2.49\times10^{-68}$

Part B:

The gradient direction is the direction of fastest increase. The direction that the temperature increase at the fastest rate is at the direction of the gradient vector.
From part A, the gradient vector is given by:
$\nabla T=200e^{-165}\left\ \textless \ -8,10,-72\right\ \textgreater \$
Thus, the direction with the fastest rate is given by:

$\frac{1}{||\left\ \textless \ -8,10,-72\right\ \textgreater \ ||} \left\ \textless \ -8,10,-72\right\ \textgreater \ \\ \\ =\frac{1}{\sqrt{(-8)^2+10^2+(-72)^2}}} \left\ \textless \ -8,10,-72\right\ \textgreater \ \\ \\ = \frac{1}{\sqrt{5348}} \left\ \textless \ -8,10,-72\right\ \textgreater \ = \frac{1}{2\sqrt{1337}} \left\ \textless \ -8,10,-72\right\ \textgreater \ \\ \\ =\frac{1}{\sqrt{1337}} \left\ \textless \ -4,5,-36\right\ \textgreater \$

Part C:

The magnitude of the maximum rate of increase is the length of the gradient vector. This is given by
$200e^{-165}\left(\sqrt{(-8)^2+10^2+(-72)^2}\right)=200\sqrt{5348}e^{-165} \\ \\ =100\sqrt{1337}e^{-165}$

8 0

3 years ago

In ⊙O, ST and VT are tangents. m∠STV = 22°. Find the value of a, b, and m∠SOV.

Rasek [7]

Answer:

$\huge \orange {\boxed {a =202\degree}}$

$\huge \purple {\boxed { b = 158\degree}}$

$\huge \red {\boxed {m\angle SOV = 158\degree}}$

Step-by-step explanation:

In $\odot$ O, ST and VT are tangents at points S and V respectively.

$\therefore OS\perp ST, \:and\: OV\perp VT$

$\therefore m\angle OST=m\angle OVT = 90\degree$

In quadrilateral OSTV,

$m\angle SOV +m\angle OST+m\angle OVT+m\angle STV = 360\degree$

(By interior angle sum postulate of a quadrilateral)

$m\angle SOV +90\degree +90\degree +22\degree = 360\degree$

$m\angle SOV +202\degree = 360\degree$

$m\angle SOV = 360\degree-202\degree$

$\huge \red {\boxed {m\angle SOV = 158\degree}}$

$\because b = m\angle SOV$

(Measure of minor arc is equal to measure of its corresponding central angle)

$\huge \purple {\boxed {\therefore b = 158\degree}}$

$\because a + b= 360\degree$

(By arc sum property of a circle)

$\therefore a = 360\degree - b$

$\therefore a = 360\degree -158\degree$

$\huge \orange {\boxed {\therefore a =202\degree}}$

6 0

3 years ago

Given f(x)=3^(x-2) and g(x)=f(3x)+4, write the function rule for function g and describe the types of transformations that occur

Liono4ka [1.6K]

Answer:

The function g(x) is $3^{3x - 2}$ + 4 and this is Translation Transformation .

Step-by-step explanation:

Given as :

The two functions f(x) and g(x) is as given

Function f(x) = $3^{x - 2}$

And g(x) = f(3 x) + 4

Now, for x = 3 x

The function f(x) can be written as

f(3 x) = $3^{3x - 2}$

So, g(x) = $3^{3x - 2}$ + 4

<u>While plotting the function f(x) and g(x) on the graph , the points translate from one to other quadrant , so this is a type of TRANSLATION TRANSFORMATION . </u>

Hence The function g(x) is $3^{3x - 2}$ + 4 and this is Translation Transformation . Answer

8 0

4 years ago

Read 2 more answers

Joe's Ice Cream Shop has 7/8 of a gallon of vanilla ice cream left. They use 1/12 of a gallon of ice cream to make one milkshake

Vadim26 [7]

Answer:

Ten and a Half milkshakes can be prepared from the available amount of ice cream.

Step-by-step explanation:

The total vanilla ice cream = 7/8 gallons

Ice cream used for 1 milkshake = 1/12 gallon

Now, total number of milkshakes possible = $\frac{\textrm{Total availibility of ice cream}}{\textrm{Ice cream used for each milkshake}}$

= $\frac{\frac{7}{8} }{\frac{1}{12} } = \frac{7}{8} \times \frac{12}{1} = 10.5$

Hence ten and a half milkshakes can be prepared from the available amount of ice cream.

8 0

4 years ago

Read 2 more answers