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3 years ago

(-9.5) to the power of 0

Mathematics

1 answer:

lukranit [14]3 years ago

5 0

Whenever it’s to the power of 0 the answer is 1 but idk if it’s negative use photomath

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ANSWER THIS ASAP DONT SEND A FILE AT ALL !! I NEED THE ANSWER QUICK HELPP ONLY ANSKER IF YOY REALLY KNOW!!

andre [41]

Answer:

...

Step-by-step explanation:

(x+8)(y+3)

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3 years ago

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Based on corresponding angles and vertical angles, which angles must always be congruent to the angles glven?

Gre4nikov [31]

Answer:

When two lines intersect they form two pairs of opposite angles, A + C and B + D. Another word for opposite angles are vertical angles. Vertical angles are always congruent, which means that they are equal. Adjacent angles are angles that come out of the same vertex

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3 years ago

Please answer these questions asap please!!! Thank you!

Len [333]

Answer:

a= 1%

b= 0.1%

c= 0.1875% probably have to round

Step-by-step explanation:

a= 2/2= 1

B= 1/10=0.1

c= 1.5/8= 0.1875

6 0

3 years ago

Determine whether the given vectors are orthogonal, parallel or neither. (a) u=[-3,9,6], v=[4,-12,-8,], (b) u=[1,-1,2] v=[2,-1,1

nevsk [136]

Answer:

a) $u v= (-3)*(4) + (9)*(-12)+ (6)*(-8)=-168$

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

$|u|= \sqrt{(-3)^2 +(9)^2 +(6)^2}=\sqrt{126}$

$|v| =\sqrt{(4)^2 +(-12)^2 +(-8)^2}=\sqrt{224}$

$cos \theta = \frac{uv}{|u| |v|}$

$\theta = cos^{-1} (\frac{uv}{|u| |v|})$

If we replace we got:

$\theta = cos^{-1} (\frac{-168}{\sqrt{126} \sqrt{224}})=cos^{-1} (-1) = \pi$

Since the angle between the two vectors is 180 degrees we can conclude that are parallel

b) $u v= (1)*(2) + (-1)*(-1)+ (2)*(1)=5$

$|u|= \sqrt{(1)^2 +(-1)^2 +(2)^2}=\sqrt{6}$

$|v| =\sqrt{(2)^2 +(-1)^2 +(1)^2}=\sqrt{6}$

$cos \theta = \frac{uv}{|u| |v|}$

$\theta = cos^{-1} (\frac{uv}{|u| |v|})$

$\theta = cos^{-1} (\frac{5}{\sqrt{6} \sqrt{6}})=cos^{-1} (\frac{5}{6}) = 33.557$

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

c) $u v= (a)*(-b) + (b)*(a)+ (c)*(0)=-ab +ba +0 = -ab+ab =0$

Since the dot product is equal to zero then the two vectors are orthogonal.

Step-by-step explanation:

For each 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.

Part a

u=[-3,9,6], v=[4,-12,-8,]

The dot product on this case is:

$u v= (-3)*(4) + (9)*(-12)+ (6)*(-8)=-168$

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:

$|u|= \sqrt{(-3)^2 +(9)^2 +(6)^2}=\sqrt{126}$

$|v| =\sqrt{(4)^2 +(-12)^2 +(-8)^2}=\sqrt{224}$

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

$cos \theta = \frac{uv}{|u| |v|}$

And the angle is given by:

$\theta = cos^{-1} (\frac{uv}{|u| |v|})$

If we replace we got:

$\theta = cos^{-1} (\frac{-168}{\sqrt{126} \sqrt{224}})=cos^{-1} (-1) = \pi$

Since the angle between the two vectors is 180 degrees we can conclude that are parallel

Part b

u=[1,-1,2] v=[2,-1,1]

The dot product on this case is:

$u v= (1)*(2) + (-1)*(-1)+ (2)*(1)=5$

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:

$|u|= \sqrt{(1)^2 +(-1)^2 +(2)^2}=\sqrt{6}$

$|v| =\sqrt{(2)^2 +(-1)^2 +(1)^2}=\sqrt{6}$

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

$cos \theta = \frac{uv}{|u| |v|}$

And the angle is given by:

$\theta = cos^{-1} (\frac{uv}{|u| |v|})$

If we replace we got:

$\theta = cos^{-1} (\frac{5}{\sqrt{6} \sqrt{6}})=cos^{-1} (\frac{5}{6}) = 33.557$

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

Part c

u=[a,b,c] v=[-b,a,0]

The dot product on this case is:

$u v= (a)*(-b) + (b)*(a)+ (c)*(0)=-ab +ba +0 = -ab+ab =0$

Since the dot product is equal to zero then the two vectors are orthogonal.

5 0

3 years ago

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In △EKL, m∠K = 90º, m∠E = 25º, EK = 3 cm, KH - altitude. Find EH.

laiz [17]

Answer:

EH = 3.31

Step-by-step explanation:

We have been given a right angle triangle EKL. As KH has been given as the altitude (perpendicular) of the right angled triangle, and K is the right angle, we can say that EK is tthe base of the triangle and EH is the only side lleft, which is the hypotenuse of the triangle.

Where,

EK = Base = 3

KH = perpendicular altitude

EH = Hypotenuse

m<K = 90

m<E = 25

We know that

cosθ = Base/ Hypotenuse

cos 25 = 3/ EH

EH = 3/cos25

EH = 3.31

Perpendicular alitutude can also be calculated by using the formula for tanθ.

3 0

2 years ago