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3 years ago

how do you use equations to solve problems about area of rectangles parallelograms trapezoids and triangles

Mathematics

1 answer:

Bogdan [553]3 years ago

8 0

Area of parellelogram and rectangle=bh

trapezoid A=(b1+b2)h÷2

triangle A=bh÷2

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Reptile [31]

Answer:

The real zeros of f(x) are x = 0.3 and x = -3.3.

Step-by-step explanation:

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$ .

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})$ , given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}$

$\bigtriangleup = b^{2} - 4ac$

In this problem, we have that:

$f(x) = x^{2} + 3x - 1$

$a = 1, b = 3, c = -1$

$\bigtriangleup = 3^{2} - 4*1*(-1) = 13$

$x_{1} = \frac{-3 + \sqrt{13}}{2*1} = 0.3$

$x_{2} = \frac{-3 - \sqrt{13}}{2*1} = -3.3$

The real zeros of f(x) are x = 0.3 and x = -3.3.

7 0

3 years ago

What is the formula for the area of a rectangle?

PtichkaEL [24]

Answer:

The area of a rectangle is length*width

Step-by-step explanation:

The answer is A

This is simple

6 0

3 years ago

Read 2 more answers

Find the solution of the initial value problem dy/dx=(-2x+y)^2-7 ,y(0)=0

Leokris [45]

Substitute $v(x)=-2x+y(x)$ , so that $\dfrac{\mathrm dv}{\mathrm dx}=-2+\dfrac{\mathrm dy}{\mathrm dx}$ . Then the ODE is equivalent to

$\dfrac{\mathrm dv}{\mathrm dx}+2=v^2-7$

which is separable as

$\dfrac{\mathrm dv}{v^2-9}=\mathrm dx$

Split the left side into partial fractions,

$\dfrac1{v^2-9}=\dfrac16\left(\dfrac1{v-3}-\dfrac1{v+3}\right)$

so that integrating both sides is trivial and we get

$\dfrac{\ln|v-3|-\ln|v+3|}6=x+C$

$\ln\left|\dfrac{v-3}{v+3}\right|=6x+C$

$\dfrac{v-3}{v+3}=Ce^{6x}$

$\dfrac{v+3-6}{v+3}=1-\dfrac6{v+3}=Ce^{6x}$

$\dfrac6{v+3}=1-Ce^{6x}$

$v=\dfrac6{1-Ce^{6x}}-3$

$-2x+y=\dfrac6{1-Ce^{6x}}-3$

$y=2x+\dfrac6{1-Ce^{6x}}-3$

Given the initial condition $y(0)=0$ , we find

$0=\dfrac6{1-C}-3\implies C=-1$

so that the ODE has the particular solution,

$\boxed{y=2x+\dfrac6{1+e^{6x}}-3}$

5 0

3 years ago

When two computers are working together, they can finish updating software in 9 minutes. How long would they take individually t

tensa zangetsu [6.8K]

First computer takes x min. For 1 min it does 1/x part of work.

Second computer takes (x+24) min. For one min it does 1/(x+24) part of work.

Both computers for one min do (1/x+1/(x+24)) part of work.

At the same time, both computers for one min do 1/9 part of work.

$\frac{1}{x} + \frac{1}{x+24} = \frac{1}{9} 
\\ \\ \frac{x+24+x}{x(x+24)} = \frac{1}{9} 
\\ \\9(2x+24)=x^{2} +24x
\\ \\ 18x+216=x^{2} +24x
\\ \\ x^{2}+6x-216=0
\\ \\ D=b^{2}-4ac=36+4*216 = 900
\\ \\ x= \frac{-b+/- \sqrt{D} }{2a} = \frac{-6+/-30}{2} 
\\ \\ x_{1}=-18, x_{2}=12

We can use only positive number here. So, 

for the first computer x=12 min,

 for second computer x+24=12+24= 36 min

Answer 12 min and 36 min.$

7 0

3 years ago

Find x please 10.3 9.6 84 9

valentinak56 [21]

The 2 small black lines through the sides of the triangle indicate that the 2 sides are the same length.

Because the 2 sides are the same the 2 bottom angles are also the same.

This means that 9x +3 = 84

Now we can solve for x:

9x +3 = 84

Subtract 3 from each side:

9x = 91

Divide both sides by 9:

x = 81 / 9

x = 9

4 0

3 years ago