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3 years ago

PLEASE HELP ASAP can you help me please

Mathematics

1 answer:

SpyIntel [72]3 years ago

8 0

So x is equal to 40 so the length of M and N is 140 and no is 70 with the total length being 210 hope I helped could I get a brainliest!!

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What’s the slope of the line?

Aleksandr [31]

Answer: 2

Step-by-step explanation:

$\frac{rise}{run} = \frac{2}{1}$

8 0

2 years ago

A group of friends were working on a student film with a budget of $700, which they

Westkost [7]

Answer:

76%

Step-by-step explanation:

First do 700 - 168 which is 532

Next we see that 700 is 100% in this scenario

Then we will use the value we seek as ‘x’

So, x% = 532

We can put the two equations, 100%/x% = 700/532

they both have LHS

We can take the inverse from both sides yields

Therefore, the answer is 76%

8 0

2 years ago

PLEASE I NEED HELP GIVING BIG POINTS!!!!

dsp73

RS over BC = RT over BA ... so RS over BC = (28/8) = (7/2) ..... RS = 10*7/2 =35 inch

8 0

3 years ago

Read 2 more answers

LK is tangent to circle J at point K. Circle J is shown. Line segment J K is a radius. Line segment K L is a tangent that inters

Evgen [1.6K]

Answer:

$(B)r=\dfrac{85}{12}$

Step-by-step explanation:

Given a circle centre J

Let the radius of the circle =r

LK is tangent to circle J at point K

From the diagram attached

LX=6
Radius, XJ=JK=r
LK=11

Theorem: The angle between a tangent and a radius is 90 degrees.

By the theorem above, Triangle JLK forms a right triangle with LJ as the hypotenuse.

Using Pythagoras Theorem:

$(6+r)^2=r^2+11^2\\(6+r)(6+r)=r^2+121\\36+6r+6r+r^2=r^2+121\\12r=121-36\\12r=85\\r=\dfrac{85}{12}$

The length of the radius, $r=\dfrac{85}{12}$

8 0

3 years ago

Read 2 more answers

Zx and Zy are supplementary angles. Zy measures 108°,

gulaghasi [49]

Answer:

72°

Step-by-step explanation:

Given

Zy = 108°

Zx = ?

We know

Zx + Zy = 180° {being supplementary angles }

Zx + 108° = 180°

Zx = 180° - 108°

Zx = 72°

Hope it will help :)

4 0

2 years ago

PLEASE HELP ASAP can you help me please​

PLEASE HELP ASAP can you help me please