All categories

2 years ago

How to use the missing pieces strategy to compare two fractions. Include a diagram with explanation

1 answer:

8 0

First u get a fraction strip or make one then go to the denominator u count by how many the nomerator is then go to the other fraction and the same thing which ever is more is the greatest hope this helped.

You might be interested in

What equation has a solution of 2/3 for x

Ahat [919]

Answer:

Step-by-step explanation:

Hope this helped! Please tell me if I'm wrong!

6 0

2 years ago

The sum of a number and 6 is greater than or equal to –4 =

Vinvika [58]

Answer:

x is greater than or = to 8

8 0

3 years ago

4. Simplify the expression:

Natalija [7]

Answer:

Step-by-step explanation:

$7x^2-6+4x+7-5x^2+x\qquad\text{combine like terms}\\\\=(7x^2-5x^2)+(4x+x)+(-6+7)\\\\=2x^2+5x+1$

3 0

3 years ago

Y = 7/8u + 9, solve for u

vivado [14]

Answer:

$u=\frac{8y-72}{7}$

Step-by-step explanation:

Flip the equation.

$\frac{7}{8}u+9=y$

Add -9 to both sides

$\frac{7}{8}u+9(-9)=y(-9)$

$\frac{7}{8}u=y-9$

Divide both sides by 7/8

$u=\frac{8y-72}{7}$

hope this helps

3 0

3 years ago

A curve is given by y=(x-a)√(x-b) for x≥b, where a and b are constants, cuts the x axis at A where x=b+1. Show that the gradient

ankoles [38]

<u>Answer:</u>

A curve is given by y=(x-a)√(x-b) for x≥b. The gradient of the curve at A is 1.

<u>Solution:</u>

We need to show that the gradient of the curve at A is 1

Here given that ,

$y=(x-a) \sqrt{(x-b)}$ --- equation 1

Also, according to question at point A (b+1,0)

So curve at point A will, put the value of x and y

$0=(b+1-a) \sqrt{(b+1-b)}$

0=b+1-c --- equation 2

According to multiple rule of Differentiation,

$y^{\prime}=u^{\prime} y+y^{\prime} u$

so, we get

${u}^{\prime}=1$

$v^{\prime}=\frac{1}{2} \sqrt{(x-b)}$

$y^{\prime}=1 \times \sqrt{(x-b)}+(x-a) \times \frac{1}{2} \sqrt{(x-b)}$

By putting value of point A and putting value of eq 2 we get

$y^{\prime}=\sqrt{(b+1-b)}+(b+1-a) \times \frac{1}{2} \sqrt{(b+1-b)}$

$y^{\prime}=\frac{d y}{d x}=1$

Hence proved that the gradient of the curve at A is 1.

7 0

2 years ago