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3 years ago

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What is the length of VW?

1 answer:

andreyandreev [35.5K]3 years ago

4 0

Answer:

2,4 is the wi5h ofe ithink

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What is tan(x)-5=0<br> (With steps)

gizmo_the_mogwai [7]

$\tan x-5=0$
$\tan x=5$

Now before taking the inverse tangent of both sides, recall that

$\tan x=\tan(x\pm\pi)=\tan(x\pm2\pi)=\cdots=\tan(x\pm n\pi)$

where $n$ is any integer. This means that when taking the inverse tangent of $\tan x$ , instead of just $x$ , we would end up with

$\arctan(\tan x)=\arctan(\tan(x+n\pi))=x\pm n\pi$

So after doing so in the equation above, we're left with

$x\pm n\pi=\arctan5\implies x=\arctan5\pm n\pi$

As $n$ can be any integer, including negative integers, we can simply write

$x=\arctan5+n\pi,\,n\in\mathbb Z$

8 0

4 years ago

A restaurant offers a $12 dinner special that has 4 choices for an appetizer, 10 choices for an entrée, and 3 choices for a d

deff fn [24]

Answer:

120 different outcomes

Step-by-step explanation:

Since there are four appetizers, ten entrees, and three desserts to choose; multiplying all the choices together will give you the total outcomes possible

3 0

3 years ago

Every week, Hector works 20 hours and earns $210.00. He earns a constant amount of money per hour.

baherus [9]

The answer is 40.

H - hours W - weeks
20 x 2 = 40

5 0

3 years ago

PLEASE PLEASE PLEASE HELP ME

lubasha [3.4K]

Answer:

CE = 17

Step-by-step explanation:

∵ m∠D = 90

∵ DK ⊥ CE

∴ m∠KDE = m∠KCD⇒Complement angles to angle CDK

In the two Δ KDE and KCD:

∵ m∠KDE = m∠KCD

∵ m∠DKE = m∠CKD

∵ DK is a common side

∴ Δ KDE is similar to ΔKCD

∴ $\frac{KD}{KC}=\frac{DE}{CD}=\frac{KE}{KD}$

∵ DE : CD = 5 : 3

∴ $\frac{KD}{KC}=\frac{5}{3}$

∴ KD = 5/3 KC

∵ KE = KC + 8

∵ $\frac{KE}{KD}=\frac{5}{3}$

∴ $\frac{KC+8}{\frac{5}{3}KC }=\frac{5}{3}$

∴ $KC + 8 = \frac{25}{9}KC$

∴ $\frac{25}{9}KC - KC=8$

∴ $\frac{16}{9}KC=8$

∴ KC = (8 × 9) ÷ 16 = 4.5

∴ KE = 8 + 4.5 = 12.5

∴ CE = 12.5 + 4.5 = 17

7 0

3 years ago

Find the area of quadrilateral ABCD whose vertices are A(1,1) B(7,-3) C(12,2) D(7,21)

sasho [114]

Answer:

The area of quadrilateral ABCD is 139 unit^2.

Step-by-step explanation:

Given:

Quadrilateral ABCD whose vertices are A(1,1) B(7,-3) C(12,2) D(7,21).

Now, to find the area.

The coordinates of the quadrilateral are A(1,1), B(7,-3), C(12,2), D(7,21).

So, to find the area we need to bisect the quadrilateral ABCD and get the triangles ABC and ADC and then calculate their areas:

In Δ ABC:

$A(x_1,y_1)=(1,1)\:,\:B(x_2,y_2)=(7,-3)\:and\:C(x_3,y_3)=(12,2)$

Now, to get the area of triangle ABC:

$Area\,of\,triangle\,=\,\frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|$

$Area\,of\,triangle\,=\,\frac{1}{2}\left|1(-3-2)+(7)(2-1)+12(1--3)\right|$

$Area\,of\,triangle\,=\,\frac{1}{2}\left|1(-5)+(7)(1)+12(4)\right|$

On solving we get:

$Area\,of\,triangle\,=25.$

In Δ ADC:

$A(x_1,y_1)=(1,1)\:,\:D(x_2,y_2)=(7,21)\:and\:C(x_3,y_3)=(12,2)$

Now, to get the area of triangle ADC:

$Area\,of\,triangle\,=\,\frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|$

$Area\,of\,triangle\,=\,\frac{1}{2}\left|1(21-2)+(7)(2-1)+12(1-21)\right|$

On solving it by same process as above we get:

$Area\,of\,triangle\,=114$

Now, to get the area of the quadrilateral we add the areas of the triangles ABC and ADC:

$25+114\\=25+114\\=139\ unit^2$

Therefore, area of quadrilateral ABCD is 139 unit^2.

4 0

3 years ago