All categories

3 years ago

(x + 40): 12 = 78 = 88 This exercise is resolved by the inverse method

Mathematics

1 answer:

pentagon [3]3 years ago

4 0

(x+40):12=88-78
(x+40):12=10
x+40=10*12
x+40=120
x=120-40
x=80

You might be interested in

Solve for x -4x-1=3 i dont know why ive been stuck on this

In-s [12.5K]

Answer:

x=-4/3

Step-by-step explanation:

x -4x-1=3

-3x-1=3

-3x-1+1=3+1

-3x=4

-3x/3 = 4/-3

x=-4/3

I hope this helps!

3 0

3 years ago

Read 2 more answers

I’m confused on this one

GenaCL600 [577]

The slope of CD is

$m = \dfrac{6 - -4}{2 - 0} = 5$

The slope of the perpendicular bisector is the negative reciprocal,

$n = -1/m = - \dfrac 1 5$

The perpendicular bisector passes through M, the midpoint of CD

M =( (0+2)/2, (-4 + 6)/2) = (1, -1)

So point slope form for the perpendicular bisector is

$y - -1 = - \frac 1 5(x - 1)$

Solving for y gives slope intercept form.

$y = - \frac 1 5 x - \frac 4 5$

Not sure how to type that without spaces; we didn't have online homework back then.

Answer: y=(-1/5)x-4/5

3 0

3 years ago

A carnival ride is in the shape of a wheel with a radius of 25 feet. The wheel has 20 cars attached to the center of the wheel.

Vladimir79 [104]

Centre angle will be 360° as its totally a circle.

radius=r=25ft

Angle between two cars=360/20=18°

Arc length=L

$\\ \sf\longmapsto L=\dfrac{\theta}{360}(2\pi r)$

$\\ \sf\longmapsto L=\dfrac{18}{360}(2\pi(25))$

$\\ \sf\longmapsto L=\dfrac{1}{20}(50\pi)$

$\\ \sf\longmapsto L=2.5\pi ft$

Now

Area be A

$\\ \sf\longmapsto A=\dfrac{1}{2}Lr^2$

$\\ \sf\longmapsto A=\dfrac{1}{2}(2.5\pi)(25)^2$

$\\ \sf\longmapsto A=625(2.5)\pi\dfrac{1}{2}$

$\\ \sf\longmapsto A=1562.5\pi/2$

$\\ \sf\longmapsto A=781\pi ft^2=$

4 0

3 years ago

$19,574 at 7.5% for 3 1/4 years

densk [106]

$1468.05...............................

6 0

3 years ago

A dentist selected 16 patients at random from a population of more than 1000 patients and asked them how many times the patient

tankabanditka [31]

The mode is the number that apppers the most on any graph.

0 appears twice
1 doesn't appear at all
2 appears once
3 appears once
4 doesn't appear at all
5 appears four times
6 appears three times
7 appears five times
8, 9, or 10 do not appear on the graph at all

So to find the mode we find the number that appears on the graph the most, the number that appears on the graph the most is 7, it appears 5 times, so 7 is the mode of this graph.

6 0

3 years ago