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fredd [130]
4 years ago
9

Prove that the altitude to the base of an isosceles triangle is also the angle bisector of the angle from which it is drawn

Mathematics
1 answer:
andriy [413]4 years ago
5 0

Step-by-step explanation:

To prove- Altitude to the base of an isosceles triangle bisects the angle from which it is drawn.

In the given figure ΔABC (isosceles triangle) whose sides AB=AC

From ∠A an altitude is drawn to base (BC) which meets BC at X

∠AXB=90 and ∠AXC =90

∴We need to prove ΔXAB= ΔXAC for altitude to bisect the ∠ BAC

In Δ AXB and Δ AXC

AB=AC (same hypotenuse)

AX= common altitude (serve as the common leg of the hypotenuse)

ΔAXB ≅ ΔAXC

Hence, we can conclude that the corresponding angle∠ XAB = ∠XAC. QED

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astraxan [27]

Answer:

  Your table is filled in below.

Step-by-step explanation:

The scale factor is the ratio of image size to original size. In part (d), the image height is "blocks" while the original height is "feet". So, the units of the scale factor are "blocks/ft".

Sometimes a scale factor has units, like this, and sometimes it is expressed as a pure number. A map scale factor might be the pure number ratio 1 : 62500, or it could also be expressed with units as 1 in : 1 mile.*

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* actually, these are slightly different scales. 1:62500 is about 0.98643 inches per mile.

7 0
3 years ago
Read 2 more answers
What is the scale factor use to create this dilation?
andrezito [222]
Given Info: "The dashed triangle is the image of the solid triangle"

So the "before" is the large solid triangle and the "after" is the smaller dashed triangle.

The horizontal side of the solid triangle is 15 units. The corresponding side for the dashed triangle is 3 units. Dividing the two values gives: 3/15 = 1/5 = 0.2

Make this value negative. The reason why is due to the fact that we have a sort of reflection going on as we're scaling down the figure. 

So the final answer is  -0.2

3 0
3 years ago
Find T5(x) : Taylor polynomial of degree 5 of the function f(x)=cos(x) at a=0 . (You need to enter function.) T5(x)= Find all va
Burka [1]

Answer:

\bf cos(x)\approx1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{4!}=\\\\=1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{24}

The polynomial is an approximation with an error less than or equals to <em>0.002652</em> for x in the interval

[-1.113826815, 1.113826815]

Step-by-step explanation:

According to Taylor's theorem

\bf f(x)=f(0)+f'(0)x+f''(0)\displaystyle\frac{x^2}{2}+f^{(3)}(0)\displaystyle\frac{x^3}{3!}+f^{(4)}(0)\displaystyle\frac{x^4}{4!}+f^{(5)}(0)\displaystyle\frac{x^5}{5!}+R_6(x)

with

\bf R_6(x)=f^{(6)}(c)\displaystyle\frac{x^6}{6!}

for some c in the interval (-x, x)

In the particular case f

<em>f(x)=cos(x) </em>

<em> </em>

we have

\bf f'(x)=-sin(x)\\f''(x)=-cos(x)\\f^{(3)}(x)=sin(x)\\f^{(4)}(x)=cos(x)\\f^{(5)}(x)=-sin(x)\\f^{(6)}(x)=-cos(x)

therefore

\bf f'(x)=-sin(0)=0\\f''(0)=-cos(0)=-1\\f^{(3)}(0)=sin(0)=0\\f^{(4)}(0)=cos(0)=1\\f^{(5)}(0)=-sin(0)=0

and the polynomial approximation of T5(x) of cos(x) would be

\bf cos(x)\approx1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{4!}=\\\\=1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{24}

In order to find all the values of x for which this approximation is within 0.002652 of the right answer, we notice that

\bf R_6(x)=-cos(c)\displaystyle\frac{x^6}{6!}

for some c in (-x,x). So

\bf |R_6(x)|\leq|\displaystyle\frac{x^6}{6!}|=\displaystyle\frac{|x|^6}{6!}

and we must find the values of x for which

\bf \displaystyle\frac{|x|^6}{6!}\leq0.002652

Working this inequality out, we find

\bf \displaystyle\frac{|x|^6}{6!}\leq0.002652\Rightarrow |x|^6\leq1.90944\Rightarrow\\\\\Rightarrow |x|\leq\sqrt[6]{1.90944}\Rightarrow |x|\leq1.113826815

Therefore the polynomial is an approximation with an error less than or equals to 0.002652 for x in the interval

[-1.113826815, 1.113826815]

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4vir4ik [10]
$78 for each tire
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The local farmers market has oranges on sale for 5 for 1.80 <br><br>What is the cost of one orange ?
lara [203]

Answer:.36 cent

1Step-by-step explanation:1.80 divided by 5

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