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4 years ago

Is ΔABC ~ ΔLMN? If so, name which similarity postulate or theorem applies.

Mathematics

1 answer:

Natasha_Volkova [10]4 years ago

4 0

Step-by-step explanation:

sss similarity theorem

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In the picture above, parallel lines p and q are cut by transversal line t. If the measure of angle 4 is 43 degrees, then the me

pshichka [43]

Answer:

angle 5 is <u>137</u> degrees

angle 1 is <u>137</u> degrees

Step-by-step explanation:

angle 5 and 6 are supplementary angles meaning together they add up to 180 degrees. to find angle 5 you must find the difference:

$180-43=137$

so angle 5 is 137 degrees.

likewise, angle 1 and angle 4 are supplementary angles. find the difference:

$180-43=137$

so angle 1 is also 137 degrees.

6 0

3 years ago

Cameron has 75% as many shirts as her brother. How many shirts does Cameron have if her brother has 12 shirts

Nataly [62]

Answer:

Step-by-step explanation:

75 percent of 12 is 9 I don't know what else to type here sorry

3 0

3 years ago

Can anyone please help : show that cos^{2} (θ)= 1 + cos(2θ) / 2 , using the compound angle formula

iragen [17]

Answer:

Below

Step-by-step explanation:

Let's prove that cos^2(O) = 1+ cos(2O)/2

We khow that cos(2O) = 2 cos^2-1

● cos(2O) = 2 cos^2-1

Add one to both sides:

● cos(2O) +1 = 2 cos^2-1+1

● cos(2O) +1 = 2 cos ^2

Divide both sides by 2

● [cos(2O)+1]/2 = cos^2

6 0

4 years ago

The model car is 3 inches long. On the box, it says the scale of the model is 1/42. What is the length of the actual car in feet

Aneli [31]

Answer:

10 feet 6 inch

Step-by-step explanation:

3 times 42 is 126. 126 to feet is 10 ft 6 inch

5 0

3 years ago

Read 2 more answers

use Taylor's Theorem with integral remainder and the mean-value theorem for integrals to deduce Taylor's Theorem with lagrange r

Vadim26 [7]

Answer:

As consequence of the Taylor theorem with integral remainder we have that

$f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots + \frac{f^{(n)}(a)}{n!}(x-a)^n + \int^a_x f^{(n+1)}(t)\frac{(x-t)^n}{n!}dt$

If we ask that $f$ has continuous $(n+1)$ th derivative we can apply the mean value theorem for integrals. Then, there exists $c$ between $a$ and $x$ such that

$\int^a_x f^{(n+1)}(t)\frac{(x-t)^k}{n!}dt = \frac{f^{(n+1)}(c)}{n!} \int^a_x (x-t)^n d t = \frac{f^{(n+1)}(c)}{n!} \frac{(x-t)^{n+1}}{n+1}\Big|_a^x$

Hence,

$\int^a_x f^{(n+1)}(t)\frac{(x-t)^k}{n!}d t = \frac{f^{(n+1)}(c)}{n!} \frac{(x-t)^{(n+1)}}{n+1} = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1} .$

Thus,

$\int^a_x f^{(n+1)}(t)\frac{(x-t)^k}{n!}d t = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}$

and the Taylor theorem with Lagrange remainder is

$f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots + \frac{f^{(n)}(a)}{n!}(x-a)^n + \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}$ .

Step-by-step explanation:

5 0

3 years ago