All categories

3 years ago

I need help ASAP!!!!!! I’m on a test! I’ll give Brainliest!

Mathematics

1 answer:

jonny [76]3 years ago

6 0

Answer:

10.6

Step-by-step explanation:

4 + 6(1.3) - 2.4 * 0.5

= 4 + 6 * 1.3 - 2.4 * 0.5

= 4 + 7.8 - 0.5 * 2.4

= 4 + 7.8 - 1.2

= 10.6

You might be interested in

A video received -16 points and 33 points in one day how many members voted

Amanda [17]

That would be 33+ (-16) so 17 members

5 0

4 years ago

Write each phrase as a numerical or algebraic expression.

Gwar [14]

4) 25 < k
5) 79 * m or 79m (depending on how your teacher taught you)
6) 325/5
7) the product of 87 and x
8) the sum of 38 and s

4 0

3 years ago

Solve the equation using the substitution u = y/x. When u = y/x is substituted into the equation, the equation becomes separable

bekas [8.4K]

Answer:

$\frac{u^2+3}{-u^3+u^2-2u}u'=\frac{1}{x}$

Step-by-step explanation:

First step: I'm going to solve our substitution for y:

$u=\frac{y}{x}$

Multiply both sides by x:

$ux=y$

Second step: Differentiate the substitution:

$u'x+u=y'$

Third step: Plug in first and second step into the given equation dy/dx=f(x,y):

$u'x+u=\frac{x(ux)+(ux)^2}{3x^2+(ux)^2}$

$u'x+u=\frac{ux^2+u^2x^2}{3x^2+u^2x^2}$

We are going to simplify what we can.

Every term in the fraction on the right hand side of equation contains a factor of $x^2$ so I'm going to divide top and bottom by $x^2$ :

$u'x+u=\frac{u+u^2}{3+u^2}$

Now I have no idea what your left hand side is suppose to look like but I'm going to keep going here:

Subtract u on both sides:

$u'x=\frac{u+u^2}{3+u^2}-u$

Find a common denominator: Multiply second term on right hand side by $\frac{3+u^2}{3+u^2}$ :

$u'x=\frac{u+u^2}{3+u^2}-\frac{u(3+u^2)}{3+u^2}$

Combine fractions while also distributing u to terms in ( ):

$u'x=\frac{u+u^2-3u-u^3}{3+u^2}$

$u'x=\frac{-u^3+u^2-2u}{3+u^2}$

Third step: I'm going to separate the variables:

Multiply both sides by the reciprocal of the right hand side fraction.

$u' \frac{3+u^2}{-u^3+u^2-2u}x=1$

Divide both sides by x:

$\frac{3+u^2}{-u^3+u^2-2u}u'=\frac{1}{x}$

Reorder the top a little of left hand side using the commutative property for addition:

$\frac{u^2+3}{-u^3+u^2-2u}u'=\frac{1}{x}$

The expression on left hand side almost matches your expression but not quite so something seems a little off.

5 0

3 years ago

#19 how do you do the work???!

Semenov [28]

Answer:

$x=\frac{1}{3}$

Step-by-step explanation:

move the terms with $x$ to the left side of the equation:

add 3x to both sides

$3x-1+3x=1$

add 3x and 3x

$6x-1=1$

move all term not containing $x$ to the right side:

add 1 to both sides of the equation

$6x=1+1$

$6x=2$

divide each term by 6

$\frac{6x}{6}=\frac{2}{6}$

$x=\frac{2}{6}$

simplify the fraction

$x=\frac{1}{3}$

8 0

3 years ago

Find the value of x and the value of y.

Zina [86]

The answer is B. x = 65, y = 10.

6 0

4 years ago