All categories

2 years ago

Three multiplied by the sum of 4 and a number is the same as 12 more than the number. Find the number.

Mathematics

1 answer:

Bess [88]2 years ago

7 0

Answer:

3(4+n) = 12+n is an equation that could be used.

12n = 12+n

11n = 12

n= 1+1/11

Step-by-step explanation:

You might be interested in

What is the value of the function at x=−2?

kogti [31]

So for this problem, the x-axis is the horizontal line in the center.

Go to where -2 is on the x-axis as shown. Use your finger to trace along since that's usually helpful in finding the point.

Move your finger from -2 on the x-axis to where the solid black line is. On the right side of that, you can see that the y-axis holds the number 2 which is what the y equals for this solid line.

Therefore, the answer should be c.) y = 2

8 0

3 years ago

B.B.ghdedfggggnmngffdgbh

Artyom0805 [142]

Just expand it, high-schooler, and what does <span>B.B.ghdedfggggnmngffdgbh mean

</span>

5 0

3 years ago

Solve 2(b − 3) ≥ 3(3 − b) and describe the graph of the solution. Show all your work.

solong [7]

Answer:

$\large\boxed{b\geq3\to b\in[3,\ \infty)}$

Step-by-step explanation:

$2(b-3)\geq3(3-b)\qquad\text{use the distributive property:}\ a(b+c)=ab+ac\\\\(2)(b)+(2)(-3)\geq(3)(3)+(3)(-b)\\\\2b-6\geq9-3b\qquad\text{add 6 to both sides}\\\\2b-6+6\geq9+6-3b\\\\2b\geq15-3b\qquad\text{add}\ 3b\ \text{to both sides}\\\\2b+3b\geq15-3b+3b\\\\5b\geq15\qquad\text{divide both sides by 5}\\\\\dfrac{5b}{5}\geq\dfrac{15}{5}\\\\b\geq3$

$,\ \geq-\text{line to the right}\\-\text{o}\text{pen circle}\\\leq,\ \geq-\text{closed circle}$

4 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

Clayton has $500 in his retirement account, and Jason has $400 in his. Clayton is adding $10 per day, whereas Jason is contribut

daser333 [38]

Answer:

$1000

Step-by-step explanation:

We can form an equation for Clayton's account: C = 500 + 10x

We can form an equation for Clayton's account: J = 400 +12x

(where x is the number of days)

When the two accounts will contain the same amount, it means: C = J

<=> 500 + 10x = 400 +12x

<=> x =50

After 50 days, there accounts will be balance. Then, we substitue x into any of the 2 equation to find out the amount: 500 + 10(50) = $1000

6 0

3 years ago