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2 years ago

Bap-fvjt-fxy aa jao sblet's talk????

Mathematics

2 answers:

Tasya [4]2 years ago

6 0

Hiii!! Do you need help in any question?

zhannawk [14.2K]2 years ago

3 0

answer: What's up?????

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Simplify the expression; 2^3 x 2^2 A. 2^5 B. 4^6 C. 4^5 D. 2^6

Illusion [34]

Answer:

Simplify the expression; 2^3 x 2^2

A. 2^5 

B. 4^6

C. 4^5

D. 2^6

Step-by-step explanation:

You add the powers together since you are multiplying.

8 0

3 years ago

Read 2 more answers

The limit as h approaches 0 of (e^(2+h)-e^2)/h is ?

Whitepunk [10]

If you plug in 0, you get the indeterminate form 0/0. You can, therefore, apply L'Hopital's Rule to get the limit as h approaches 0 of e^(2+h),
which is just e^2.
[e^(2+h) − e^2]/h = [e^2(e^h−1)]/h

so in the limit, as h goes to 0, you'll notice that the numerator and denominator each go to zero (e^h goes to 1, and so e^h-1 goes to zero). This means the form is 'indeterminate' (here, 0/0), so we may use L'Hoptial's rule:

=e^2

3 0

3 years ago

In the accompanying diagram of ABC ca is extended to D, m∠ABC = 70 and m∠BCA =50 Find m∠DAB

zvonat [6]

Answer:

120 degrees

Step-by-step explanation:

Find the diagram attached.

From the diagram;

Interior angles are m∠BCA and m∠ABC

Exterior angle is m∠DAB

The sum of interior angle of the triangle is equal to exterior

m∠BCA +m∠ABC =m∠DAB

Given

m∠ABC = 70

m∠BCA =50

m∠DAB = 70 + 50

m∠DAB = 120 degrees

Hence the measure of m∠DAB is 120 degrees

6 0

3 years ago

Find the measures of the angles of the triangle whose vertices are A = (-3,0) , B = (1,3) , and C = (1,-3).A.) The measure of ∠A

alekssr [168]

Answer:

$\theta_{CAB}=128.316$

$\theta_{ABC}=25.842$

$\theta_{BCA}=25.842$

Step-by-step explanation:

A = (-3,0) , B = (1,3) , and C = (1,-3)

We're going to use the distance formula to find the length of the sides:

$r= \sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}$

$AB= \sqrt{(-3-1)^2+(0-3)^2}=5$

$BC= \sqrt{(1-1)^2+(3-(-3))^2}=9$

$CA= \sqrt{(1-(-3))^2+(-3-0)^2}=5$

we can use the cosine law to find the angle:

it is to be noted that:

the angle CAB is opposite to the BC.

the angle ABC is opposite to the AC.

the angle BCA is opposite to the AB.

to find the CAB, we'll use:

$BC^2 = AB^2+CA^2-(AB)(CA)\cos{\theta_{CAB}}$

$\dfrac{BC^2-(AB^2+CA^2)}{-2(AB)(CA)} =\cos{\theta_{CAB}}$

$\cos{\theta_{CAB}}=\dfrac{9^2-(5^2+5^2)}{-2(5)(5)}$

$\theta_{CAB}=\arccos{-\dfrac{0.62}}$

$\theta_{CAB}=128.316$

Although we can use the same cosine law to find the other angles. but we can use sine law now too since we have one angle!

To find the angle ABC

$\dfrac{\sin{\theta_{ABC}}}{AC}=\dfrac{\sin{CAB}}{BC}$

$\sin{\theta_{ABC}}=AC\left(\dfrac{\sin{CAB}}{BC}\right)$

$\sin{\theta_{ABC}}=5\left(\dfrac{\sin{128.316}}{9}\right)$

$\theta_{ABC}=\arcsin{0.4359}\right)$

$\theta_{ABC}=25.842$

finally, we've seen that the triangle has two equal sides, AB = CA, this is an isosceles triangle. hence the angles ABC and BCA would also be the same.

$\theta_{BCA}=25.842$

this can also be checked using the fact the sum of all angles inside a triangle is 180

$\theta_{ABC}+\theta_{BCA}+\theta_{CAB}=180$