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2 years ago

Find the value of y so that the line passing through the points (2,5) and (7, y) has a slope of 2/3

Mathematics

1 answer:

borishaifa [10]2 years ago

8 0

Use the bmxy chart b should be your other ¨y¨ so do 5= 2/3(2)+b and whatever u get for ¨b¨ should be your answer

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How do u solve this question in the picture

dedylja [7]

1/2 over -2/5 + (-1/4)

10/20 over -8/20 + (-5/20)

1 5/20 + (-5/20)

1

The answer is 1

6 0

2 years ago

I got the same answer as the example

DaniilM [7]

Answer:

Step-by-step explanation:

2x2=4 4-7=3 3x3=21 21-4=17

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2 years ago

I Need Help On Multiplying Fractions It’s 5th Grade

nikdorinn [45]

Answer:

Ok ill help

Step-by-step explanation:

7 0

2 years ago

You can make a tent by throwing a piece of cloth over a clothesline, then securing the edge to the ground with stakes so that an

4vir4ik [10]

Based on the required width and height of the tent, the length of the cloth would have to be 14.4 ft.

<h3>How long should the length of the cloth be?</h3>

The opening of the tent would be an isosceles triangle but if it is divided in two from the top, a right-angled triangle will be formed and the hypotenuse will be the length of the cloth on one side.

That length - according to the Pythagoras theorem - would be:

= √(6² + (8 /2)²)

= √(36 + 16)

= 7.2 ft

As this is the length of one side of the cloth, the length of the entire cloth is:

= 7.2 x 2

= 14.4 ft

Find out more on the Pythagoras theorem at brainly.com/question/343682.

#SPJ1

8 0

1 year ago

Who can help me d e f thanks

12345 [234]

$y = (2ax^2 + c)^2 (bx^2 - cx)^{-1}$

Product rule:

$y' = \bigg((2ax^2+c)^2\bigg)' (bx^2-cx)^{-1} + (2ax^2+c)^2 \bigg((bx^2-cx)^{-1}\bigg)'$

Chain and power rules:

$y' = 2(2ax^2+c)\bigg(2ax^2+c\bigg)' (bx^2-cx)^{-1} - (2ax^2+c)^2 (bx^2-cx)^{-2} \bigg(bx^2-cx\bigg)'$

Power rule:

$y' = 2(2ax^2+c)(4ax) (bx^2-cx)^{-1} - (2ax^2+c)^2 (bx^2-cx)^{-2} (2bx - c)$

Now simplify.

$y' = \dfrac{8ax (2ax^2+c)}{bx^2 - cx} - \dfrac{(2ax^2+c)^2 (2bx-c)}{(bx^2-cx)^2}$

$y' = \dfrac{8ax (2ax^2+c) (bx^2 - cx) - (2ax^2+c)^2 (2bx-c)}{(bx^2-cx)^2}$

$y = \dfrac{3bx + ac}{\sqrt{ax}}$

Quotient rule:

$y' = \dfrac{\bigg(3bx+ac\bigg)' \sqrt{ax} - (3bx+ac) \bigg(\sqrt{ax}\bigg)'}{\left(\sqrt{ax}\right)^2}$

$y'= \dfrac{\bigg(3bx+ac\bigg)' \sqrt{ax} - (3bx+ac) \bigg(\sqrt{ax}\bigg)'}{ax}$

Power rule:

$y' = \dfrac{3b \sqrt{ax} - (3bx+ac) \left(-\frac12 \sqrt a \, x^{-1/2}\right)}{ax}$

Now simplify.

$y' = \dfrac{3b \sqrt a \, x^{1/2} + \frac{\sqrt a}2 (3bx+ac) x^{-1/2}}{ax}$

$y' = \dfrac{6bx + 3bx+ac}{2\sqrt a\, x^{3/2}}$

$y' = \dfrac{9bx+ac}{2\sqrt a\, x^{3/2}}$

$y = \sin^2(ax+b)$

Chain rule:

$y' = 2 \sin(ax+b) \bigg(\sin(ax+b)\bigg)'$

$y' = 2 \sin(ax+b) \cos(ax+b) \bigg(ax+b\bigg)'$

$y' = 2a \sin(ax+b) \cos(ax+b)$

We can further simplify this to

$y' = a \sin(2(ax+b))$

using the double angle identity for sine.

7 0

11 months ago