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3 years ago

Quadrilateral QRST is a square. If the measure of angle RQT is 3x - 6. Find the value of x.

Mathematics

1 answer:

trapecia [35]3 years ago

3 0

Given that quadrilateral QRST is a square.

Each angle of a square is of 90 degrees.

Angle <RQT is also an angle of 90 degrees.

We also given angle RQT = 3x - 6.

So, we can setup an equation as 3x-6 =90.

Now, we need to solve the equation for x.

6 is being subtracted from left side.

We always apply reverse operation. Reverse operation of subtraction is addition.

So, adding 6 on both sides of the equation, we get

3x-6+6 =90+6.

3x = 96.

3 is being multipied with x, in order to remove that 3, we need to apply reverse operation of multiplication.

So, dividing both sides by 3.

$\frac{3x}{3} = \frac{96}{3}$

x=32 (final answer).

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Please help me solve this problem ASAP

DiKsa [7]

$\bold{\huge{\blue{\underline{ Solution }}}}$

<h3>Given :-</h3>

The right angled below is formed by 3 squares A, B and C
The area of square B has an area of 144 inches ²
The area of square C has an of 169 inches ²

We have to find the area of square A?

<h3>Let's Begin :- </h3>

The right angled triangle is formed by 3 squares

We have, 

Area of square B is 144 inches²
Area of square C is 169 inches²

We know that,

$\bold{ Area \: of \: square = Side × Side }$

Let the side of square B be x

Subsitute the required values,

$\sf{ 144 = x × x }$

$\sf{ 144 = x² }$

$\sf{ x = √144}$

$\bold{\red{ x = 12\: inches }}$

Thus, The dimension of square B is 12 inches

Area of square C = 169 inches

Let the side of square C be y

Subsitute the required values,

$\sf{ 169 = y × y }$

$\sf{ 169 = y² }$

$\sf{ y = √169}$

$\bold{\green{ y = 13\: inches }}$

Thus, The dimension of square C is 13 inches.

It is mentioned in the question that, the right angled triangle is formed by 3 squares

The dimensions of square be is x and y

Let the dimensions of square A be z

<h3>Therefore, By using Pythagoras theorem, </h3>

The sum of squares of base and perpendicular height equal to the square of hypotenuse

That is,

$\bold{\pink{ (Perpendicular)² + (Base)² = (Hypotenuse)² }}$

Here, 

Base = x = 12 inches
Perpendicular = z
Hypotenuse = y = 13 inches

Subsitute the required values,

$\sf{ (z)² + (x)² = (y)² }$

$\sf{ (z)² + (12)² = (169)² }$

$\sf{ (z)² + 144 = 169}$

$\sf{ (z)² = 169 - 144 }$

$\sf{ (z)² = 25}$

$\bold{\blue{ z = 5 }}$

Thus, The dimensions of square A is 5 inches

<h3>Therefore,</h3>

Area of square

$\sf{ = Side × Side }$

$\sf{ = 5 × 5 }$

$\bold{\orange{ = 25\: inches }}$

Hence, The area of square A is 25 inches.

6 0

2 years ago

Solve the initial value problem. dy/dt = 1 + 6/t , t > 0, y = 8 when t = 1

nordsb [41]

$\displaystyle\frac{dy}{dt} = 1 + \frac{6}{t}\ \Rightarrow\ dy = \left( 1 + \frac{6}{t}\right) dt\ \Rightarrow\int 1\, dy = \int \left( 1 + \frac{6}{t}\right) dt \ \Rightarrow \\ \\
y = t + 6\ln|t| + C. \text{ But $t\ \textgreater \ 0$ so }y = t + 6\ln t + C. \\ \\ 
y(1) = 8\ \Rightarrow\ 8 = 1 + 6 \ln 1 + C \ \Rightarrow\ C = 7 \text{ so } \\ \\
y(t) = t + 6\ln t + 7$