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3 years ago

Find the missing side length image below

Mathematics

1 answer:

barxatty [35]3 years ago

4 0

Answer:

Step-by-step explanation:

Based on the Proportional Transversal Theorem, the three parallel lines hat intersects the two transversals, divides the transversal lines proportionally.

Therefore, we would have the following ratio:

28/35 = ?/50

Cross multiply

35*? = 50*28

35*? = 1,400

Divide both sides by 35

? = 1400/35

? = 40

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Find the root of each of the following equation x+8=24

kupik [55]

Answer:

Step-by-step explanation:

If you mean root of x, 24-8=16

square root of 16=4

4 0

2 years ago

You open a drink stand where you sell coffee and hot chocolate. You sell coffee for $1.80 and hot chocolate for $3.15. At the en

Usimov [2.4K]

Answer:

c = 69 and h = 51

Step-by-step explanation:

You will need two different equations here

use variable c as coffee and variable h as hot chocolate

c + h = 120

1.80c + 3.15h = 284.40

By moving variables, you can alter the first equation

h = 120 - c

Now insert this into the second equation

1.80c + 3.15(120 - c) = 284.40

Multiply

1.80c + 378 - 3.15c = 284.40

Subtract the c

-1.35c + 378 = 284.40

Subtract 378 from both sides

-1.35c = -93.6

Divide both sides by -1.35

c = 69

Then plug into the first equation

69 + h = 120

Subtract 69

h = 51

7 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

Read 2 more answers

A freezer can be bought on hire purchase by making a desposiy of 15% of the cash price which is $2975 the interest which is 20%

Liono4ka [1.6K]

Answer: See explanation

Step-by-step explanation:

a. The deposit

= 15% × $2975

= 15/100 × $2975

= 0.15 × $2975

= $446.25

b. The hire purchase price

Outstanding balance = $2975 - $446.25 = $2528.75

20% of outstanding balance

= 20% × $2528.75

= 0.2 × $2528.75

= $505.75

Hire purchase price

= $2975 + $505.75

= $3480.75

c. The difference the hire purchase price and cash price

= $3480.75 - $2975

= $505.75

d. The difference as a percentage of the cash price.

= ($505.75 / $2975) × 100

= 0.17 × 100

= 17%

5 0

2 years ago

At the beginning of October, you have $300 in your bank account. You get $25 of allowance each week. Your cousin has $200 in her

lys-0071 [83]

Answer:

20 weeks to have the same amount for both bank accounts

Step-by-step explanation:

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6 0

2 years ago