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1 year ago

A, B & C lie on a straight line.

Mathematics

2 answers:

Mama L [17]1 year ago

7 0

Answer:

Step-by-step explanation:

i dont know how to explain i just know thats the answer hahaha

Musya8 [376]1 year ago

6 0

Answer:

Step-by-step explanation:

Since triangle DBC is isosceles, angles DBC and DCB must be congruent. Angle BDC is 40 so the sum of the measures of the aforementioned angles must be 180 - 40 = 140; divide that by two and you get that the measure of angle DBC is 70. Angles DBC and ABD are supplementary, as they lie on a straight line, so we can deduce the measure of angle DBA is 180 - 70 = 110. The measure of angle DAB is 180 minus the measures of angle DBA and ADB, we figured out angle DBA is 110 degrees and the problem tells us angle ADB is 45 degrees, so we can subtract to get the measure of angle DAB to be 180 - (110 + 45) = 25.

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3 years ago

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2 years ago

A square base of a pyramid has the dimensions 5 yards by 5 yards. The height of one of the triangular faces is 12 yards. How can

-Dominant- [34]

Answer:

145 yards squared

Step-by-step explanation:

So the area of the base is area of a square with side length of 5 yards $5*5=25$

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4 0

2 years ago

Read 2 more answers

Enter the coefficients of the fifth Taylor polynomial T5(x) for the function f(x) = x5−3x4+2x2+5x−2 based at b=1. T5(x)= + (x−1)

DENIUS [597]

Compute the necessary values/derivatives of $f(x)$ at $x=1$ :

$f(1)=3$

$f'(1)=2$

$f''(1)=-12$

$f'''(1)=-12$

$f^{(4)}(1)=48$

$f^{(5)}(1)=120$

Taylor's theorem then says we can "approximate" (in quotes because the Taylor polynomial for a polynomial is another, exact polynomial) $f(x)$ at $x=1$ by

$T_5(x)=\dfrac3{0!}+\dfrac2{1!}(x-1)-\dfrac{12}{2!}(x-1)^2-\dfrac{12}{3!}(x-1)^3+\dfrac{48}{4!}(x-1)^4+\dfrac{120}{5!}(x-1)^5$