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3 years ago

Find the perimeter of the following shape, rounded to the nearest tenth:

Mathematics

2 answers:

zaharov [31]3 years ago

8 0

I think 18 :333333333

Kryger [21]3 years ago

4 0

I think the answer is 18?

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I NEED HELP ASAP!!!!!! I WILL GIVE PRAILIEST TO THE CORRECT ANSWER

Norma-Jean [14]

-13 (q + 4) = 7q + 16

I think to solve it you do this first:

-13q + -52 = 7q + 16

And then I think you do:

-6q + -36

And thats all I can remember sorry ):

That -6q + -36 might be the answer I think (:

Hope I could help (:

3 0

3 years ago

The area of a rectangular piece of cardboard is represented by

Liula [17]

This question is solved applying the formula of the area of the rectangle, and finding it's width. To do this, we solve a quadratic equation, and we get that the cardboard has a width of 1.5 feet.

Area of a rectangle:

The area of rectangle of length l and width w is given by:

$A = wl$

w(2w + 3) = 9

From this, we get that:

$l = 2w + 3, A = 9$

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:

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

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

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

In this question:

$w(2w+3) = 9$

$2w^2 + 3w - 9 = 0$

Thus a quadratic equation with $a = 2, b = 3, c = -9$

Then

$\Delta = 3^2 - 4(2)(-9) = 81$

$w_{1} = \frac{-3 + \sqrt{81}}{2*2} = 1.5$

$w_{2} = \frac{-3 - \sqrt{81}}{2*2} = -3$

Width is a positive measure, thus, the width of the cardboard is of 1.5 feet.

Another similar problem can be found at brainly.com/question/16995958

5 0

3 years ago

Identify the values of the variables. Give your answers in simplest radical form. HELP ASAP!!

kumpel [21]

Answer:

Step-by-step explanation:

Since the triangle is right use the sine/ tangent ratios to solve for x and y

note sin60° = $\frac{\sqrt{3} }{2}$ and tan60° = $\sqrt{3}$

sin60° = $\frac{opposite}{hypotenuse}$ = $\frac{3\sqrt{6} }{y}$

ysin60° = 3 $\sqrt{6}$

y × $\frac{\sqrt{3} }{2}$ = 3 $\sqrt{6}$

multiply both sides by 2 and divide by $\sqrt{3}$

y v= 6 $\sqrt{\frac{6}{3} }$ = 6 $\sqrt{2}$

tan60° = $\frac{opposite}{adjacent}$ = $\frac{3\sqrt{6} }{x}$

xtan60° = 3 $\sqrt{6}$

x × $\sqrt{3}$ = 3 $\sqrt{6}$

divide both sides by $\sqrt{3}$

x = 3 $\sqrt{\frac{6}{3} }$ = 3 $\sqrt{2}$

8 0

3 years ago

Read 2 more answers

Which quadratic equation has solutions x=8 and x=−1?

andrezito [222]

There are two ways you can do this

$Using the formula$

${x}^{2} - (sum \: of \: roots)x + poduct \: of \: roots = 0$

${x}^{2} - (8+-1)x + - 1 \times 8 = 0$

${x}^{2} - 7x - 8 = 0$

$Working backwards$

$x = 8 \: or \: x = - 1$
$\Rightarrow x-8=0 \: or \: x+1=0$

$\Rightarrow (x-8)(x + 1)=0$
$\Rightarrow {x}^{2} + x -8x - 8 = 0$
$\Rightarrow {x}^{2} - 7x - 8 = 0$

7 0

4 years ago

Prac

PtichkaEL [24]

Answer:

Yes a number with 3 digits is usually bigger than a number with 2 digits but sometimes if it is a decimal than the decimal can be bigger

8 0

3 years ago