All categories

3 years ago

Jayne uses long division to convert a rational number

Mathematics

1 answer:

VARVARA [1.3K]3 years ago

6 0

Answer:

Answer: The decimal form of a/b never repeats or terminates.

Step-by-step explanation:

Fractions such as " a/b " can be ratios so they don't terminate.

You might be interested in

Select the correct product.

AURORKA [14]

Answer:

Option A. $2x^{3}+7x^{2}+12x-8$

Step-by-step explanation:

we have

$(x^{2}+4x+8)(2x-1)$

Applying the distributive property

$(x^{2}+4x+8)(2x-1) =x^{2}(2x)-x^{2} +4x(2x)-4x+8(2x)-8\\ \\=2x^{3}-x^{2} +8x^{2}-4x+16x-8\\ \\= 2x^{3}+7x^{2}+12x-8$

7 0

3 years ago

Can someone please help me answer this question asap thank you

kykrilka [37]

Equation would be y=x+5/8

6 0

3 years ago

Read 2 more answers

Which of the following of x makes the rational expressions below undefined

alisha [4.7K]

A rational expression is undefined if the denominator is zero. Since the denominator is $x+13$ , the expression is undefined if

$x+13 = 0 \iff x=-13$

5 0

3 years ago

What is the equation for line D?

Free_Kalibri [48]

Answer:

Step-by-step explanation:

7 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

3 years ago