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podryga [215]
3 years ago
9

Write an equation to represent the following description. The difference of seven times a number and 15 is 41.

Mathematics
2 answers:
chubhunter [2.5K]3 years ago
7 0

Answer: 3

7x + 15 = 41 or 7x - 15 = 41

egoroff_w [7]3 years ago
6 0

Answer:

7x - 15 = 41

Step-by-step explanation:

The key word is "difference" so keep that in mind. 7 times a number and 15 is 41. The "number" is unknown so a variable can be put in place until the number is known. The whole equation equals 41. The word "and" is where the "difference" or subtraction sign should go. Now we have a translated description! I hope this helps and if I'm wrong, please let me know!

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What is the prime factorizarion for 500​
Nataly_w [17]

Answer:

2 exponents (2) times 5 exponents (3)

Step-by-step explanation:

To find the prime factors you start by dividing the number by the first prime number which is 2 if there is not a remainder meaning you can divide evenly anymore write down how many 2's you were able to divide by evenly now try dividing by the the next prime factor which is 3 the goal is to get to a quotient of 1

5 0
3 years ago
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Help!! Ill give 100 points!
Volgvan

Answer:

It will take about half an hour more for maria to finish his/her project than Juan.

Step-by-step explanation:

9/10 > 2/6

=> 54/60 > 20/60

=> 34/60

Therefore, it will take about half an hour more for maria to finish his/her project than Juan.

Hoped this helped.

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3 years ago
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PLEASEEEE HELP I WILL GIVE BRAINLIEST!!
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Answer:

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Step-by-step explanation:

3 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
4 years ago
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The city park in which Allen plays is a square and measures exactly one million square feet. Which side dimensions produce the m
valentina_108 [34]
A- 998.8×997.7=996502.76
B- 998.84×997.73=996572.6332
C- 998.843×997.731=996576.6252
D - 999×998=997002

So D is the answer
4 0
4 years ago
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