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3 years ago

Can somebody help me with this question ?? :)

Mathematics

2 answers:

S_A_V [24]3 years ago

8 0

Answer:

x = 44°

Step-by-step explanation:

Since this is an isosceles triangle, the other angle that isn't defined is also 68°. So, since all the angles add up to 180°,

x + 68(2) = 180

x + 136 = 180

x = 44°

scoundrel [369]3 years ago

6 0

Answer:

44 is the answer

Step-by-step explanation:

You might be interested in

find the angle between the vectors. (first find the exact expression and then approximate to the nearest degree. ) a=[1,2,-2]. B

SashulF [63]

Answer:

$\theta = cos^{-1} (\frac{10}{\sqrt{9} \sqrt{25}})=cos^{-1} (\frac{10}{15}) = cos^{-1} (\frac{2}{3}) = 48.190$

Since the angle between the two vectors is not 180 or 0 degrees we can conclude that are not parallel

And the anfle is approximately $\theta \approx 48$

Step-by-step explanation:

For this case first we need to calculate the dot product of the vectors, and after this if the dot product is not equal to 0 we can calculate the angle between the two vectors in order to see if there are parallel or not.

a=[1,2,-2], b=[4,0,-3,]

The dot product on this case is:

$a b= (1)*(4) + (2)*(0)+ (-2)*(-3)=10$

Since the dot product is not equal to zero then the two vectors are not orthogonal.

Now we can calculate the magnitude of each vector like this:

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

$|b| =\sqrt{(4)^2 +(0)^2 +(-3)^2}=\sqrt{25}= 5$

And finally we can calculate the angle between the vectors like this:

$cos \theta = \frac{ab}{|a| |b|}$

And the angle is given by:

$\theta = cos^{-1} (\frac{ab}{|a| |b|})$

If we replace we got:

$\theta = cos^{-1} (\frac{10}{\sqrt{9} \sqrt{25}})=cos^{-1} (\frac{10}{15}) = cos^{-1} (\frac{2}{3}) = 48.190$

Since the angle between the two vectors is not 180 or 0 degrees we can conclude that are not parallel

And the anfle is approximately $\theta \approx 48$

3 0

2 years ago

Guys please help! In triangle CDE, the measure of angle C is 90 degrees, side CD is 15 and DE is 3 more than CE. Find the length

AURORKA [14]

You can calculate it using the law of cosines: c^2=a^2+b^2-2*a*b*cos(C)

your triangle is
CD=15=a
CE=?=b
DE=CE+3=b+3=c
and C=90°

-> insert those values, with c substituted with b+3 to remove c

c^2=a^2+b^2-2*a*b*cos(C)
(b+3)^2=15^2+b^2-2*15*b*cos(90)
cos(90)=0->
(b+3)^2=15^2+b^2
b^2+2*3*b+3^2=225+b^2
6b+9=225
6b=216
b=36=CE

DE=CE+3=36+3=39

4 0

3 years ago

Solve for x, show all your work: <br> x3 = 10

Alex73 [517]

3.3333333333333333333
repeating

8 0

3 years ago

Read 2 more answers

Which of the following represents the additive inverse of h?

erica [24]

Subtract ............

6 0