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Alona [7]
2 years ago
10

Need help!

Mathematics
1 answer:
siniylev [52]2 years ago
8 0

<u>Correct </u><u>Inputs </u><u>:-</u>

In ΔABC right angled at A, D and E are points on BC, C such that BD = CD and AD ⊥ BC

\underline{\underline{\large\bf{Solution:-}}}\\

\longrightarrow Let us know about definition of altitude first. The altitude of a triangle is the perpendicular line segment drawn from the vertex to the opposite side of the triangle.

\leadstoMedian is the line segment from a vertex to the midpoint of the opposite side.

<u>Let us Check all options one by one </u>

  • CD is line segment which starts from vertex C but don't falls on opposite side AB thus it is not an altitude.❌

  • BA is line segment which starts from vertex B and falls perpendicularly on opposite sides AC and is thus an altitude.✔️

  • AD is line segment which starts from vertex A and falls perpendicularly on opposite side BC and is thus an altitude.✔️

  • AE is a line segment which starts from vertex A but doesn't falls perpendicularly on opposite side BC and is thus not an altitude.❌

  • AD falls on BC with D as mid point because BD = CD and is thus a median. ✔️
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deff fn [24]
The answer would be:
-7sqrt(6x)/4x^3
6 0
3 years ago
Solve for w. 9w-4/4=5w-6/7+6​
ELEN [110]

Answer:

w=\frac{1}{16}

Step-by-step explanation:

\frac{9w-4}{4} =\frac{5w-6}{7+6}

9w-1=\frac{5w-6}{13}

9w-1+1=\frac{5w-6}{13}+1

9w=\frac{5w-6}{13}+1<u />

13 × 9w = 13 × \frac{5w-6}{13}+1

117w = 5w - 6 + 13

117w = 5w + 7

117w - 5w = 5w - 5w + 7

112w = 7

112w ÷ 112 = 7 ÷ 112

w=\frac{1}{16}

5 0
3 years ago
Consider the curve defined by the equation y=6x2+14x. Set up an integral that represents the length of curve from the point (−2,
torisob [31]

Answer:

32.66 units

Step-by-step explanation:

We are given that

y=6x^2+14x

Point A=(-2,-4) and point B=(1,20)

Differentiate w.r. t x

\frac{dy}{dx}=12x+14

We know that length of curve

s=\int_{a}^{b}\sqrt{1+(\frac{dy}{dx})^2}dx

We have a=-2 and b=1

Using the formula

Length of curve=s=\int_{-2}^{1}\sqrt{1+(12x+14)^2}dx

Using substitution method

Substitute t=12x+14

Differentiate w.r t. x

dt=12dx

dx=\frac{1}{12}dt

Length of curve=s=\frac{1}{12}\int_{-2}^{1}\sqrt{1+t^2}dt

We know that

\sqrt{x^2+a^2}dx=\frac{x\sqrt {x^2+a^2}}{2}+\frac{1}{2}\ln(x+\sqrt {x^2+a^2})+C

By using the formula

Length of curve=s=\frac{1}{12}[\frac{t}{2}\sqrt{1+t^2}+\frac{1}{2}ln(t+\sqrt{1+t^2})]^{1}_{-2}

Length of curve=s=\frac{1}{12}[\frac{12x+14}{2}\sqrt{1+(12x+14)^2}+\frac{1}{2}ln(12x+14+\sqrt{1+(12x+14)^2})]^{1}_{-2}

Length of curve=s=\frac{1}{12}(\frac{(12+14)\sqrt{1+(26)^2}}{2}+\frac{1}{2}ln(26+\sqrt{1+(26)^2})-\frac{12(-2)+14}{2}\sqrt{1+(-10)^2}-\frac{1}{2}ln(-10+\sqrt{1+(-10)^2})

Length of curve=s=\frac{1}{12}(13\sqrt{677}+\frac{1}{2}ln(26+\sqrt{677})+5\sqrt{101}-\frac{1}{2}ln(-10+\sqrt{101})

Length of curve=s=32.66

5 0
3 years ago
Which of the following expressions is NOT a perfect square trinomial?
alisha [4.7K]

Answer:

2

Step-by-step explanation:

4 0
3 years ago
The manager at a movie theater keeps track of the number of evening tickets and matinee tickets sold each day and the total mone
Alex Ar [27]
Let the number of Matinee tickets be x 
Let the number of evening tickets be y
Number of sold tickets = 63, this means that: x+y = 63 ......> equation I
Price of matinee ticket = 10 and price of evening tickets = 14
Total money collected = 742, this means that 10x + 14y = 742 .....> equation II

From equation I: x = 63 - y .........> equation III

Substitute by equation III in equation II:
10(63-y) + 14y = 742
630 - 10y + 14y = 742
4y = 742 - 630 = 112
y = 112/4 = 28
Substitute by the value of y is equation III:
x = 63-28 = 35

Based on the above calculations:
number of matinee tickets = x = 35 tickets
number of evening tickets = y = 28 tickets
4 0
3 years ago
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