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2 years ago

ANSWER ASAP!!!!!!!!!!!!!!!!

Mathematics

1 answer:

djverab [1.8K]2 years ago

3 0

Answer:

all of the above

Step-by-step explanation:

hope this helps.

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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]

Answer:

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

Solution:

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.

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2 years ago

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nexus9112 [7]

Answer:

Its C i think

Step-by-step explanation:

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3 years ago

Which of these strategies would eliminate a variable in the system of equations?

KATRIN_1 [288]

Answer: A and B

Explanation: I’ll use A as an example. When you multiply the bottom equation by -3/2, the equation becomes: -9x+6y=-3. Then, proceed to add the equations. Look at X and Y in both equations: do the Y variables in both equations equal to 0 when added together? Yes, because -6y+6y=0! Therefore, you can eliminate y in these two equations.

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