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Arisa [49]
3 years ago
14

What is the answer to -3(x+4)+15=6-4x

Mathematics
1 answer:
vredina [299]3 years ago
5 0

The answer is -12

Remember pemdas

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Step-by-step explanation:

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Graph the line.<br>y - 4x = -8​
stiv31 [10]
X = -8-y/4 hope this helps
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A customer can drive around a track 60 times for $12. At this rate, how many times can the customer drive around the track for $
Aleonysh [2.5K]

Answer:36 times for 8 dollars

Step-by-step explanation:

k so first you divide 54 and 12 =4.5 per dollar so you then multiply 4.5 and 8 to get the answer that =36 times for 8 dollars

6 0
3 years ago
Read 2 more answers
FIVE STARS AND BRAINLIEST FOR CORRECT ANSWER
Charra [1.4K]

For the given vectors \vec{a}=a_{1}\hat{i}+a_{2}\hat{j} and \vec{b}=b_{1}\hat{i}+b_{2}\hat{j}

The dot product of vectors a and b is defined as = \vec{a}.\vec{b}=(a_{1}\times b_{1}+a_{2} \times b_{2})

So, \vec{a}.\vec{b} = (4\hat{i}+3\hat{j}). (-4\hat{i}+4\hat{j})

\vec{a}.\vec{b}=(4\times(-4)+3 \times4)

= -16+12

= -4

8 0
3 years ago
Let φ(x, y) = arctan (y/x) .
Alexandra [31]

Answer:

a) \large F(x,y)=(-\frac{y}{x^2+y^2},\frac{x}{x^2+y^2})

b) \large \mathbb{R}^2-\{(0,0)\}

c) the points of the form (x, -x) for x≠0

Step-by-step explanation:

a)

If φ(x, y) = arctan (y/x), the vector field F = ∇φ would be

\large F(x,y)=(\frac{\partial \phi}{\partial x},\frac{\partial \phi}{\partial y})

On one hand we have,

\large \frac{\partial \phi}{\partial x}=\frac{\partial arctan(y/x)}{\partial x}=\frac{-y/x^2}{1+(y/x)^2}=-\frac{y/x^2}{1+y^2/x^2}=\\\\=-\frac{y/x^2}{(x^2+y^2)/x^2}=-\frac{y}{x^2+y^2}

On the other hand,

\large \frac{\partial \phi}{\partial y}=\frac{\partial arctan(y/x)}{\partial y}=\frac{1/x}{1+(y/x)^2}=\frac{1/x}{1+y^2/x^2}=\\\\=\frac{1/x}{(x^2+y^2)/x^2}=\frac{x}{x^2+y^2}

So

\large F(x,y)=(-\frac{y}{x^2+y^2},\frac{x}{x^2+y^2})

b)

The domain of definition of F is  

\large \mathbb{R}^2-\{(0,0)\}

i.e., all the plane X-Y except the (0,0)

c)

Here we want to find all the points such that

\large (-\frac{y}{x^2+y^2},\frac{x}{x^2+y^2})=(k,k)

where k is a real number other than 0.

But this means

\large -\frac{y}{x^2+y^2}=\frac{x}{x^2+y^2}\Rightarrow y=-x

So, all the points in the line y = -x except (0,0) are parallel to the vector field F, that is, the points (x, -x) with x≠ 0

8 0
3 years ago
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