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Mariulka [41]
3 years ago
9

Does the equation 2x + 3 = 11 go through the point (1,4)?

Mathematics
1 answer:
Jet001 [13]3 years ago
6 0
Answer: yes it does

2x= 8
x= 8/2

x=4.
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10 POINTS PLEASE HELP
Llana [10]

Answer:

A company sells tents in two styles, shown below.

The sides and floor of each tent made of nylon.

For tunnel tent:

In a composite figure of tunnel tent :

There are two semi circle, half cylindrical shape and its base is made of Square.

Area of semi-circle is given by:

where r is the radius the circle.

Here, Diameter(d) = 8 feet

We know that: Diameter(d) = 2 r

then;

8 = 2r

⇒ r= 4 ft.

Area of square =

then;

Area of square =  square ft

Lateral surface area of cylinder =  where r is the radius and h is the height

Here, h = 8 ft

Lateral surface area of half cylinder =  square ft

Total Surface area of tunnel tent(S)=2(Area of semi-circle)+(Lateral surface area of half cylindrical)+(Area of square)

Substitute the given values we have;

square ft.

For Pup tent:

Each side of tent = 8 ft

In a figure of Pup tent:

There are 3 square shape and two triangle.

Area of square =

then;

Area of square =  square ft

Area of a triangle is given by:

Each side of a triangle is same.

This is an equilateral triangle, and its height is given by:

then;

Area of triangle = square ft.

Substitute the given values we have;

Total surface area of Pup tent(S')

=  square ft.

⇒S' > S

therefore, the tunnel tent require less nylon to manufacture.

<h2 /><h2>PLZ GIMME BRAINLIEST :)) !!!!!!</h2>

6 0
3 years ago
Pleaseee helppp with thissss asapppp
Natasha2012 [34]

As we know that the standard equation of circle is {\bf{(x-h)^{2}+(y-k)^{2}=r^{2}}} , where <em>r</em> is the radius of circle and centre at <em>(h,k) </em>

Now , as the circle passes through <em>(2,9)</em> so it must satisfy the above equation after putting the values of <em>h</em> and <em>k</em> respectively

{:\implies \quad \sf \{2-(-1)\}^{2}+(9-5)^{2}=r^{2}}

{:\implies \quad \sf (3)^{2}+(4)^{2}=r^{2}}

{:\implies \quad \sf 9+16=r^{2}}

After raising ½ power to both sides , we will get <em>r = +5 , -5</em> , but as radius can never be -<em>ve</em> . So <em>r = +</em><em>5</em><em> </em>

Now , putting values in our standard equation ;

{:\implies \quad \sf \{x-(-1)\}^{2}+(y-5)^{2}=(5)^{2}}

{:\implies \quad \bf \therefore \quad \underline{\underline{(x+1)^{2}+(y-5)^{2}=25}}}

<em>This is the required equation of </em><em>Circle</em>

Refer to the attachment as well !

8 0
2 years ago
Use the Law of Sines to solve the triangle. (Let b = 47.7 yd. Round your answers for a and c to two decimal places.)
klio [65]

Answer:

C = 68.667°

a = 123.31 yd.

c = 114.90 yd.

Step-by-step explanation:

The missing image for the question is attached to this solution.

In the missing image, a triangle AB is given with angles A and B given to be 88° 35' and 22° 45' respectively

We are them told to find angle C and side a and c given that side b = 47.7 yd.

A = 88° 35' = 88° + (35/60)° = 88.583°

B = 22° 45' = 22° + (45/60)° = 22.75°

The sum of angles in a triangle = 180°

A + B + C = 180°

C = 180° - (A + B) = 180° - (88.583° + 22.75°) = 68.667°

The sine law is given as

(a/sin A) = (b/sin B) = (c/sin C)

Using the first two terms of the sine law

(a/sin A) = (b/sin B)

a = ?

A = 88.583°

b = 47.7 yd.

B = 22.75°

(a/sin 88.583°) = (47.7/sin 22.75°)

a = (47.7 × sin 88.583°) ÷ sin 22.75°

a = 123.31 yd.

Using the last two terms of the sine law

(b/sin B) = (c/sin C)

b = 47.7 yd.

B = 22.75°

c = ?

C = 68.667°

(47.7/sin 22.75°) = (c/sin 68.667°)

c = (47.7 × sin 68.667°) ÷ sin 22.75°

c = 114.90 yd.

Hope this Helps!!!

8 0
3 years ago
2 The diagram shows a
Strike441 [17]

Diagram? There is no diagram given.

a.

i =  \frac{180(n - 2)}{n}

i =  \frac{180(12 - 2)}{12}  =  \frac{180(10)}{12}  =  \frac{1800}{12}  \\ i = 150\degree

Cannot solve b without the diagram.

5 0
3 years ago
Some one please help me this is my final question i think i have the right answers but im not sure
Xelga [282]
Asked and answered elsewhere.
brainly.com/question/9436435
3 0
3 years ago
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