All categories

3 years ago

JK has endpoints at J(5,9) and K(7, 7). Find the midpoint M of JK.

Mathematics

2 answers:

Basile [38]3 years ago

5 0

Answer:

J (7, 10)

K (3, 2)

Formula for midpoint:

x - coordinate for M:

y - coordinate for M:

Therefore, M (5,6).

Hope this helps :)

Step-by-step explanation:

Aleksandr-060686 [28]3 years ago

5 0

Answer:

$(6,8 )$

Step-by-step explanation:

Midpoint Formula: $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2} )$

Simply plug in your coordinates into the midpoint formula to find midpoint:

$(\frac{5+7}{2},\frac{9+7}{2} )$

$(\frac{12}{2},\frac{16}{2} )$

$(6,8 )$

You might be interested in

I need the answer for a= and b=

Lerok [7]

The answer is 67 and 34!

3 0

3 years ago

MULTIPLE CHOICE:

Vanyuwa [196]

Answer:

4.7 and 5.1

Step-by-step explanation:

7.4 only works if it shows less than or equal to inequality

5 0

3 years ago

F(x) = 4x + 7, g(x) = 3x^2 Find (f + g)(x)

kondor19780726 [428]

For the functions f(x) = 4x + 7 and g(x) = -9x + 11, find f(g(0)). For the functions f(x) = 3x2 - 7 and g(x) = -3x2 + 4, find f(g(2)). (Do NOT put commas in large numbers.)

3 0

3 years ago

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

Divide 14 by v. Then, subtract 7.

Nastasia [14]

Answer:

14-7v

--------

Step-by-step explanation:

i dont understand what else the answer could be, due to the fact that not enough information if provided to result in a solid sum.

3 0

3 years ago