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3 years ago

What landmarks are located in East India?

Mathematics

1 answer:

Setler79 [48]3 years ago

7 0

Answer:

Mehrangarh Fort.

Amber Fort. Located on the outskirts of Jaipur, the Amber Fort (also known as the Amer Fort) is a landmark of phenomenal beauty. ...

Jaisalmer Fort.

Step-by-step explanation:

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Help pls pls pls !!!!

kramer

Answer:

AAS

Step-by-step explanation:

I believe

6 0

3 years ago

Enter an equation for the function that includes the points.Give your answer in a(b)x. In the event that a=1 , give your answer

Andrews [41]

Answer:

$f(x) = \frac{24}{25} * \frac{5}{6}^x$

Step-by-step explanation:

Given

$(x_1,y_1) = (2,\frac{2}{3})$

$(x_2,y_2) = (3,\frac{5}{9})$

Required

Write the equation of the function $f(x) = ab^x$

Express the function as:

$y = ab^x$

In: $(x_1,y_1) = (2,\frac{2}{3})$

$y = ab^x$

$\frac{2}{3} = a * b^2$ --- (1)

In $(x_2,y_2) = (3,\frac{5}{9})$

$y = ab^x$

$\frac{5}{9} = a * b^3$ --- (2)

Divide (2) by (1)

$\frac{5}{9}/\frac{2}{3} = \frac{a*b^3}{a*b^2}$

$\frac{5}{9}/\frac{2}{3} = b$

$\frac{5}{9}*\frac{3}{2} = b$

$\frac{5}{3}*\frac{1}{2} = b$

$\frac{5}{6} = b$

$b = \frac{5}{6}$

Substitute 5/6 for b in (1)

$\frac{2}{3} = a * b^2$

$\frac{2}{3} = a * \frac{5}{6}^2$

$\frac{2}{3} = a * \frac{25}{36}$

$a = \frac{2}{3} * \frac{36}{25}$

$a = \frac{2}{1} * \frac{12}{25}$

$a = \frac{24}{25}$

The function: $f(x) = ab^x$

$f(x) = \frac{24}{25} * \frac{5}{6}^x$

7 0

3 years ago

Please answer question three and four if you can :)<br> Show full working out ty;)

nikklg [1K]

Step-by-step explanation:

Let D be the mid point of side BC, [B(2, - 1), C(5, 2)].

Therefore, by mid-point formula:

$D = ( \frac{2 + 5}{2}, \: \: \frac{ - 1 + 2}{2} ) = ( \frac{7}{2}, \: \: \frac{ 1}{2} ) \\ \therefore D= (3.5, \: \: 0.5) \\ \& \: A=(-1,\:\:4)...(given) \\\\ now \: by \: distance \: formula \\ \\ Length \: of \: segment \: AD \\ = \sqrt{( - 1 - 3.5)^{2} + {(4 - 0.5)}^{2} } \\ = \sqrt{(4.5)^{2} + {(3.5)}^{2} } \\ = \sqrt{20.25 + 12.25 } \\ = \sqrt{32.5} \\ \red{ \boxed{\therefore Length \: of \: segment \: AD = 5.7 \: units}}$

4 (a)

Equation of line AB[A(2, 1), B(-2, - 11)] in two point form is given as:

$\frac{y-y_1}{y_1-y_2} =\frac{x-x_1}{x_1 - x_2} \\\\\therefore \frac{y-1}{1-(-11)} =\frac{x-2}{2 - (-2) } \\\\\therefore \frac{y-1}{1+11} =\frac{x-2}{2 +2} \\\\\therefore \frac{y-1}{12} =\frac{x-2}{4} \\\\\therefore \frac{y-1}{3} =\frac{x-2}{1} \\\\\therefore y-1= 3(x - 2)\\\\\therefore y= 3x - 6+1\\\\\therefore y= 3x - 5\\\\ \huge \purple {\boxed {\therefore 3x - y-5=0}} \\$

is the equation of line AB.

Now we have to check whether C(4, 7) lie on line AB or not.

Let us substitute x = 4 & y = 7 on the Left hand side of equation of line AB and if it gives us 0, then C lies on the line.

$LHS = 3x - y-5\\=3\times 4-7-5\\= 12-12\\=0\\= RHS$

Hence, point C (4, 7) lie on the straight line AB.

4(b)

Like we did in 4(a), first find the equation of line AB and then substitute the coordinates of point C in equation and if they satisfy the equation, then all the three points lie on the straight line.

4 0

4 years ago

Some body please help me!!!!!!

VladimirAG [237]

Surface area of a cube = 486 square inches

Solution:

Given each side of a cube = 9 inches

Net of a cube has 6 squares.

Area of square = side × side

Area of 1 square = 9 × 9 = 81 square inches

Surface area of a cube = Area of 6 squares

= 6 × (side × side)

= 6 × 81

= 486 square inches

Hence, surface area of a cube is 486 square inches.

7 0

3 years ago

PLEASE HELP ME WITH THIS!! ILL GIVE BRAINLIEST.

LekaFEV [45]

Answer:

See below.

Step-by-step explanation:

A correct.

B correct.

C correct.

D correct.

E incorrect.

F correct.

G incorrect.

H correct.

I incorrect.

J incorrect.

K correctt.

-hope it helps

4 0

2 years ago