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3 years ago

Write the coordinates

Mathematics

1 answer:

Ilia_Sergeevich [38]3 years ago

6 0

What kind of coordinates?

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I think the parent function is

y=x

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3 years ago

Find the gradient of the line segment between the points (10,-4) and (6,-16).

svet-max [94.6K]

The answer is -3 . Use the gradient formula y2-y1/x2-x1

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2 years ago

Solve the two step equations

disa [49]

Hello!

1) -3x - 4 = 23

-3x = 23 + 4

-3x = 27

x = 27 : (-3)

x = -9

2) x/2 - 12 = -4

x - 24 = -8

x = -8 + 24

x = 16

3) 6a + (-1) = 10

6a - 1 = 10

6a = 10 + 1

6a = 11

a = 11 : 6

a = 11/6

4) -(x + 2) = 12

x + 2 = -12

x = -12 - 2

x = -14

5) 7a + 12 = 10

7a = 10 - 12

7a = -2

a = -2 : 7

a = -2/7

6) -4(a + 2) = 12

a + 2 = -3

a = -3 - 2

a = -5

Good luck! :)

7 0

3 years ago

What is the percentage of sugar in the syrup made of 10kg of water and 5kg of sugar? Hi and have a good day :D

arsen [322]

Answer:

33.33%

Step-by-step explanation:

the syrup has a total of 15kg of ingredients, 5kg of it is sugar, so 15/5 which will end as 33.33%

8 0

3 years ago

Read 2 more answers

A student is given that point P(a, b) lies on the terminal ray of angle Theta, which is between StartFraction 3 pi Over 2 EndFra

Harman [31]

Answer:

A.

The student made an error in step 3 because a is positive in Quadrant IV; therefore,

$cos\theta = \frac{a\sqrt{a^2 + b^2}}{a^2 + b^2}$

Step-by-step explanation:

Given

$P\ (a,b)$

$r = \± \sqrt{(a)^2 + (b)^2}$

$cos\theta = \frac{-a}{\sqrt{a^2 + b^2}} = -\frac{\sqrt{a^2 + b^2}}{a^2 + b^2}$

Required

Where and which error did the student make

Given that the angle is in the 4th quadrant;

The value of r is positive, a is positive but b is negative;

Hence;

$r = \sqrt{(a)^2 + (b)^2}$

Since a belongs to the x axis and b belongs to the y axis;

$cos\theta$ is calculated as thus

$cos\theta = \frac{a}{r}$

Substitute $r = \sqrt{(a)^2 + (b)^2}$

$cos\theta = \frac{a}{\sqrt{(a)^2 + (b)^2}}$

$cos\theta = \frac{a}{\sqrt{a^2 + b^2}}$

Rationalize the denominator

$cos\theta = \frac{a}{\sqrt{a^2 + b^2}} * \frac{\sqrt{a^2 + b^2}}{\sqrt{a^2 + b^2}}$

$cos\theta = \frac{a\sqrt{a^2 + b^2}}{a^2 + b^2}$

So, from the list of given options;

The student's mistake is that a is positive in quadrant iv and his error is in step 3

3 0

3 years ago

Write the coordinates​

Write the coordinates