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3 years ago

1/15.999=x/54.5 Does anyone know what to do? I keep getting the answer wrong.

2 answers:

8 0

Hello!

First, let's write the problem.
$\frac{1}{15.999}=\frac{x}{54.5}$
Let's switch sides now, otherwise, we are going to end up doing it later.
$\frac{x}{54.5}=\frac{1}{15.999}$
Multiply both sides by 54.5.
$\frac{x}{54.5} *54.5 =\frac{1}{15.999}*54.5$
We could also rewrite that as,
$\frac{54.5x}{54.5}=\frac{1\cdot \:54.5}{15.999}$
Simplify.
$\frac{54.5x}{54.5}=x$
$\frac{1\cdot \:54.5}{15.999}=\frac{54.5}{15.999}$
$x=3.40646\dots$
That would be our final answer, we could round it to.
x ≈ 3.40

You can feel free to let me know if you have any questions regarding this!
Thanks!

- TetraFish

Degger [83]3 years ago

3 0

Hi Ballentinesavannah I'm going to show u my work of how I got the answer.

0.062504 = x/54.5
0.062504 * 54.5 = x
3.406463 = x
switch it

Answer: x = 3.406463

In order to get the answer I will explain it to u step by step : simplify 1/15.999 to 0.062504, multiply both sides by 54.5, then multiply 0.062504 * 54.5 so it can equal 3.406463.

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4 years ago

I need help ASAP. Thank you :)

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m∠FDE = 52°

Solution:

Given data:

DE ≅ DF, CD || BE, BC || FD and m∠ABF = 116°

Sum of the adjacent angles in a straight line = 180°

m∠ABF + m∠CBF = 180°

116° + m∠CBF = 180°

m∠CBF = 64°

If CD || BE, then CD || BF.

Hence CD || BE and BE || FD.

Therefore BFCD is a parallelogam.

In parallelogram, Adjacent angles form a linear pair.

m∠CBF + m∠BFD = 180°

64° + m∠BFD = 180°

m∠BFD = 116°

Sum of the adjacent angles in a straight line = 180°

m∠BFD + m∠DFE = 180°

116° + m∠DFE = 180°

m∠DFE = 64°

we know that DE ≅ DF.

In triangle, angles opposite to equal sides are equal.

m∠DFE = m∠DEF

m∠DEF = 64°

sum of all the angles of a triangle = 180°

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3 years ago

Find the area of the trapezoid below to the nearest tenth.

kondor19780726 [428]

Answer:

Area of trapezoid = 67.6 square units

Step-by-step explanation:

Area of a trapezoid is given by the expression,

Area of the trapezoid = $\frac{1}{2}(b_1+b_2)h$

Here, $b_1$ and $b_2$ are the parallel sides of the given trapezoid.

And ' $h$ ' = Height between the parallel sides

From the given triangle ABE,

m(∠ABE) = m(∠ABC) - m(∠EBC)

m(∠ABE) = 120° - 90°

= 30°

By applying cosine rule in the given triangle,

cos(30°) = $\frac{\text{Adjacent side}}{\text{Hypotenuse}}$

$\frac{\sqrt{3} }{2}=\frac{BE}{AB}$

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

BE = $3\sqrt{3}$ units

By applying sine rule in ΔABE,

sin(30°) = $\frac{\text{Opposite side}}{\text{Hypotenuse}}$

$\frac{1}{2}=\frac{AE}{AB}$

$\frac{1}{2}=\frac{AE}{6}$

AE = 3 units

Length of $b_1=BC=10$

Length of $b_2=AD=(AE+EF+FD)$ [AE = FD, since given trapezoid ABCD is an isosceles trapezoid]

$b_2=3+10+3$

$b_2=16$

Height between the parallel sides $h=3\sqrt{3}$

Area of the trapezoid = $\frac{1}{2}(BC+AD)BE$

= $\frac{1}{2}(10+16)(3\sqrt{3})$

= $39\sqrt{3}$

= 67.6 square units

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3 years ago

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