3 years ago

The Difference of 4 and k, divided by 16.

Mathematics

1 answer:

miskamm [114]3 years ago

5 0

Answer:4-k÷6

Step-by-step explanation:

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3 years ago

7y+3=3x+11 find x and y

kvasek [131]

Answer:

7y + 3 = 3x + 11

7y = 3x + 11 - 3

7y = 3x + 8

y = 3x + 8 / 7

7y + 3 = 3x + 11

7y + 3 - 11 = 3x

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3 years ago

What would m equal if -6.85+m/4=-11

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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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3 years ago

Store B sells the same computer as in the Example for $1,300. Store B offers a 20% discount on the computer. The tax rate is the

I am Lyosha [343]

Answer:

$1,300 discount of 20%=$260

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2 years ago