All categories

4 years ago

What fraction would you add to 29/6 to make 4

1 answer:

8 0

Keeping in mind that 29/6 is greater than 4, is actually 4 and 5/6, so the amount we'll "add" will be a negative one.

$\bf \cfrac{29}{6}+x=4\implies x=4-\cfrac{29}{6}\implies x=\cfrac{4}{1}-\cfrac{29}{6}\implies x=\cfrac{(6)4-(1)29}{6}
\\\\\\
x=\cfrac{24-29}{6}\implies x=\cfrac{-5}{6}\\\\
-------------------------------\\\\
\cfrac{29}{6}+\left( -\cfrac{5}{6} \right)=4$

You might be interested in

The product of 7 and the sum of a number and -6

Hoochie [10]

This is middle school work woah

5 0

3 years ago

A school principal ordered 1,000 pencils. He gave an equal number to each of 7 teachers until he had given out as many as possib

Semenov [28]

1000/7 = <span>142.857142857

Because this is not an even number we take "142", then times it by 7 teachers.

142*7= 994

1000 - 994 = 6 Pencils left</span>

3 0

4 years ago

Solve for w. 7(-2w + 3) = 1 - 4w <br>what is the value of w?

blondinia [14]

Answer:

w=2

Step-by-step explanation:

-14w + 21 = 1 - 4w

-10w + 21 = 1

-10w = -20

w = -20/-10

w=2

5 0

3 years ago

Read 2 more answers

Janet is 3 years older than her sister Julia Janice brother is 8 years old younger than their sister Julia the sum of all of the

Allushta [10]

Answer: It would be 11 years old

Step-by-step explanation: uJst in case you still

4 0

3 years ago

Solve the differential. This was in the 2016 VCE Specialist Maths Paper 1 and i'm a bit stuck

Nimfa-mama [501]

$\sqrt{2 - x^{2}} \cdot \frac{dy}{dx} = \frac{1}{2 - y}$
$\frac{dy}{dx} = \frac{1}{(2 - y)\sqrt{2 - x^{2}}}$

Now, isolate the variables, so you can integrate.
$(2 - y)dy = \frac{dx}{\sqrt{2 - x^{2}}}$
$\int (2 - y)\,dy = \int\frac{dx}{\sqrt{2 - x^{2}}}$
$2y - \frac{y^{2}}{2} = sin^{-1}\frac{x}{\sqrt{2}} + \frac{1}{2}C$

$4y - y^{2} = 2sin^{-1}\frac{x}{\sqrt{2}} + C$
$y^{2} - 4y = -2sin^{-1}\frac{x}{\sqrt{2}} - C$
$(y - 2)^{2} - 4 = -2sin^{-1}\frac{x}{\sqrt{2}} - C$
$(y - 2)^{2} = 4 - 2sin^{-1}\frac{x}{\sqrt{2}} - C$

$y - 2 = \pm\sqrt{4 - 2sin^{-1}\frac{x}{\sqrt{2}} - C}$
$y = 2 \pm\sqrt{4 - 2sin^{-1}\frac{x}{\sqrt{2}} - C}$

At x = 1, y = 0.
$0 = 2 \pm\sqrt{4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C}$
$-2 = \pm\sqrt{4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C}$

$4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C > 0$
$\therefore 2 = \sqrt{4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C}$

$4 = 4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C$
$0 = -2sin^{-1}\frac{1}{\sqrt{2}} - C$
$C = -2sin^{-1}\frac{1}{\sqrt{2}} = -2\frac{\pi}{4}$
$C = -\frac{\pi}{2}$

$\therefore y = 2 - \sqrt{4 + \frac{\pi}{2} - 2sin^{-1}\frac{x}{\sqrt{2}}}$

6 0

4 years ago