Answer:
The correct answer is – 4/7
The tangent line to <em>y</em> = <em>f(x)</em> at a point (<em>a</em>, <em>f(a)</em> ) has slope d<em>y</em>/d<em>x</em> at <em>x</em> = <em>a</em>. So first compute the derivative:
<em>y</em> = <em>x</em>² - 9<em>x</em> → d<em>y</em>/d<em>x</em> = 2<em>x</em> - 9
When <em>x</em> = 4, the function takes on a value of
<em>y</em> = 4² - 9•4 = -20
and the derivative is
d<em>y</em>/d<em>x</em> (4) = 2•4 - 9 = -1
Then use the point-slope formula to get the equation of the tangent line:
<em>y</em> - (-20) = -1 (<em>x</em> - 4)
<em>y</em> + 20 = -<em>x</em> + 4
<em>y</em> = -<em>x</em> - 24
The normal line is perpendicular to the tangent, so its slope is -1/(-1) = 1. It passes through the same point, so its equation is
<em>y</em> - (-20) = 1 (<em>x</em> - 4)
<em>y</em> + 20 = <em>x</em> - 4
<em>y</em> = <em>x</em> - 24
You multiple the area 18 x 12 = 96 sq ft
For the first question, divide 32 by 40 and it should equal .8 miles. Then, for the second question, multiply .8 by 13 and it should equal 10.4 miles.
Answer: A. "Segment AD bisects angle CAB." is the right answer.
Step-by-step explanation:
Given : In ΔABC ,AC≅AB.
⇒∠ACB=∠CBA....(1) (∵ angles opposite to equal sides of a triangle are equal )
Now in ΔACD and ΔABD
AD=AD (common)....(2)
Here we need one more statement to prove the triangles congruent that is only statement (A) fits in it.
If AD bisects ∠CAB then ∠CAD=∠BAD..(3)
Now again Now in ΔACD and ΔABD
∠ACB=∠CBA [from (1)]
AD=AD [common]
∠CAD=∠BAD [from (3)]
So by ASA congruency criteria ΔADC≅ΔABD.