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3 years ago

What is the value of the expression (e+c-d)x dn when e=-6, n=2, c=6, and d=-1

Mathematics

1 answer:

saul85 [17]3 years ago

8 0

The value of the expression (e+c-d)x dn is -2

The given expression is:

(e+c-d)x dn

Substitute e=-6, n=2, c=6, and d=-1 into (e + c - d) x dn

(e+c-d)x dn = [-6 + 6 -(-1)] x (-1)(2)

(e+c-d)x dn = (0 + 1) x (-2)

(e+c-d)x dn = 1 x (-2)

(e+c-d)x dn = -2

Learn more here: brainly.com/question/13961297

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Answer:

3 3/5

Step-by-step explanation:

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2 years ago

The graph shows the number of hours per day spent on social media by a group of teenagers and the number of hours per day spent

umka21 [38]

Answer:

0.9 hours

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Step-by-step explanation:

The attached graph is the drawn graph of given data

i) A teenager would spend exercising in one day if they spent 0.25 hours on social media.

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According to the graph if you locate the 0.25 hour on x-axis and draw perpendicular the will intersect the line on the value of y-axis is = 0.9 hours

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4 years ago

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il63 [147K]

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8 0

3 years ago

I came up the answer as 57. I will attach my note, can you check?

ololo11 [35]

BC=19

Explanation

Step 1

ABE

triangle ABE is rigth triangle, then let

$\begin{gathered} Angle=60 \\ adjacentside=BE \\ opposit\text{ side(the one in front of the angle)= AB=}\frac{19\sqrt[]{6}}{4} \end{gathered}$

so, we need a function that relates, angle, adjancent side and opposite side

$\tan \theta=\frac{opposite\text{ side}}{\text{adjacent side}}$

replace

$\begin{gathered} \tan \theta=\frac{opposite\text{ side}}{\text{adjacent side}} \\ \tan 60=\frac{AB}{\text{BE}} \\ \text{cross multiply} \\ \text{BE}\cdot\tan \text{ 60=AB} \\ \text{divide both sides by tan 60} \\ \frac{\text{BE}\cdot\tan\text{ 60}}{\tan\text{ 60}}=\frac{\text{AB}}{\tan\text{ 60}} \\ BE=\frac{\text{AB}}{\tan\text{ 60}} \\ \text{if AB=}\frac{19\sqrt[]{6}}{4} \\ BE=\frac{\frac{19\sqrt[]{6}}{4}}{\sqrt[]{3}} \\ BE=\frac{19\sqrt[]{6}}{4\sqrt[]{3}} \end{gathered}$

Step 2

BED

again, we have a rigth triangle,then let

$\begin{gathered} \text{Hypotenuse}=BD \\ \text{adjacent side= BE=6.71} \\ \text{angle}=\text{ 45} \end{gathered}$

so, we need a function that relates; angle, hypotenuse and adjacent side

$\cos \theta=\frac{adjacent\text{ side}}{\text{hypotenuse}}$

replace.

$\begin{gathered} \cos \theta=\frac{adjacent\text{ side}}{\text{hypotenuse}} \\ \cos 45=\frac{6.71}{\text{BD}} \\ BD=\frac{6.71}{\cos \text{ 45}} \\ BD=\frac{\frac{19\sqrt[]{6}}{4\sqrt[]{3}}}{\frac{\sqrt[]{2}}{2}} \\ BD=\frac{38\sqrt[]{6}}{4\sqrt[]{6}} \\ BD=\frac{38}{4} \end{gathered}$

Step 3

finally BDE

let

angle=30

opposite side= BD

use sin function

$\begin{gathered} \sin \theta=\frac{opposite\text{ side}}{\text{hypotenuse}} \\ \text{replace} \\ \sin \text{ 30=}\frac{BD}{BC} \\ BC\cdot\sin 30=BD \\ BC=\frac{BD}{\sin \text{ 30}} \\ BC=\frac{\frac{38}{4}}{\frac{1}{2}} \\ BC=\frac{76}{4}=19 \\ BC=19 \end{gathered}$

so, the answer is 19

I hop

4 0

1 year ago

Find the least common multiple of the following polynomials: 5y^2-80 and y+4

algol [13]

$5y^2 - 80 = 5(y^2-16) = 5(y - 4)(y + 4)$

LCM of 5(y - 4)(y + 4) and (y + 4) is 5(y - 4)(y + 4)

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Answer: 5(y - 4)(y + 4) (Answer D)
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3 0

3 years ago

Read 2 more answers