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3 years ago

Help me please!!!!!!!!

Mathematics

2 answers:

yanalaym [24]3 years ago

7 0

Answer:

Step-by-step explanation:

X=5 and 3

ella [17]3 years ago

5 0

Answer: a. x=3 or 5

Step-by-step explanation:

x^2-8x+15=0

When factoring this equation we need to find 2 numbers that multiply to get positive (15) and add to get negative (-8)

So -3 and -5 fit this description because -3+-5=-8 and -3*-5=15

(x-3)(x-5)=0

Solve for values of x to get x=3 and x=5

so X=3 or 5

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never [62]

0.75 (1/2x) = 60

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So her total remaining money is $80.

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3 years ago

There is a bag with only red marbles and blue marbles. The probability of randomly choosing a red marble is 2 9 . There are 63 m

Tatiana [17]

The main idea is the number of marbles and the denominator of the probability. All you have to do is multiply the fraction by a fraction which is equal to 1 which makes the denominator equal to 63. We know 63 is a multiple of 9. 9 goes 7 times in 63 so we multiple the fraction $\frac{2}{9}$ with $\frac{7}{7}$ which gives, $\frac{2 * 7}{9 * 7}$ which when evaluated gives $\frac{14}{63}$ and the numerator is the number of red marbles in the bag. So there are 14 red marbles in the bag :D

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3 years ago

Long division by single digit <br>

Sauron [17]

Answer:

Here's an example of long division:

Step-by-step explanation:

Let's just do 48 divided by 6. First you have to multiply the 6 (the number we're dividing with) until it reaches the closest to 48 (6 x 8 = 48). Now you subtract 48 from 48, and you get zero. So 48 divided by 6 would be 0.

Hope this answers your question!

7 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:

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,

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