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2 years ago

Guys help me pls help me

Mathematics

1 answer:

Semenov [28]2 years ago

4 0

Answer:

10 units

Step-by-step explanation:

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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

Which number could be added to

Arturiano [62]

The number which could be added to both sides of the equation to complete the square is; -2.25.

<h3>Which number could be added to both sides of this quadratic equation to complete the square?</h3>

The quadratic equations give in the task content is; 1=x²-3x. Hence, to express the equation by completing the square method; we have;

1 - 2.25 = (x -3/2)² - 2.25

Hence, the number which should be added to both sides of the equation is; -2.25.

Read more on completing the square;

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1 year ago

What is one way the Pennsylvania colony was able to maintain peace with the Native American people?

myrzilka [38]

B. Hshejsiaiajsgaiajbehsiabsvahajakajbsjs

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2 years ago

Kate made tables of values to solve a system of equations. First she found that the x-value of the solutionwas between 0 and 1,

Blizzard [7]

Answer:

D. (0.6, 1.3)

Step-by-step explanation:

The difference between y-values is smallest for x=0.6. The approximate y-value is reasonably chosen as the average of the y-values for that value of x.

(x, y) = (0.6, 1.3) is a reasonable approximation