All categories

3 years ago

Solve the following inequality. 6.37 <-45.36 Which graph shows the correct solution?

Mathematics

1 answer:

ale4655 [162]3 years ago

6 0

Answer:

third number line

Step-by-step explanation:

6.3r < -45.36

r < -7.2

it can't be the first two because those values are 71, 70 etc and we need 7.1, 7.2 etc

and r is less then -7.2

So it has to be the third one

You might be interested in

Given the following diagram, enter the required information.

KonstantinChe [14]

M(CXA) = m(BXA) + m(CXB)
= 30°20' + 40°35'
m(CXA) = 70°55'

6 0

3 years ago

Read 2 more answers

I will give brainliest to whoever answers this question. Also if you can please explain how you solved it I would appreciate it!

LUCKY_DIMON [66]

The answer is 4.9. 32/26=6/x, cross multiple to get 32x=156, divide by 32 to get x=4.875~4.9

3 0

3 years ago

Read 2 more answers

If x = 10, y = 5, and z = 11, what is the sine ratio for g°?

salantis [7]

Answer:

$\frac{10}{11}$

Step-by-step explanation:

<u><em>The picture of the question in the attached figure</em></u>

we know that

In the right triangle ABC

$sin(g^o)=\frac{AC}{AB}$ ----> by SOH (opposite side divided by the hypotenuse)

substitute the given values

$sin(g^o)=\frac{x}{z}$

$sin(g^o)=\frac{10}{11}$

8 0

2 years ago

At noon, ship A is 130 km west of ship B. Ship A is sailing east at 25 km/h and ship B is sailing north at 20 km/h. How fast is

Hitman42 [59]

Answer:

answer = 12.87 km/h

Step-by-step explanation:

Given

Ship A is sailing east at 25 km/h = $\frac{dx}{dt}$

ship B is sailing north at 20 km/h = $\frac{dy}{dt}$

here x and y are the sailing at t = 4 : 00 pm for ship A and B respectively

so we get x = 4 ×25 =100 km/h

y = 4× 20 = 80 km/h

let z is the distance between the ships, we need to find $\frac{dz}{dt}$ at t = 4 hr

At noon, ship A is 130 km west of ship B (12:00 pm)

so equation will be

$z^2 = (130-x)^2 + y^2......................(i)\\put x = 100 and y = 80 \\\\we | | get \\$

$z^2 = 30^2 + 80^2\\z =\sqrt{7300} km/h$

derivative first equation w . r. to t we get

$2z\frac{dz}{dt} =$ $-2(130-x)\frac{dx}{dt}$ $+2y\frac{dy}{dt}$

$\frac{dz}{dt} =\frac{1}{z}[(x -130)\frac{dx}{dt} +y\frac{dy}{dt}]$

$\frac{dz}{dt} = \frac{( -20\times25 + 80\times20)}{\sqrt{7300} }$

$= \frac{1100}{85.44}\\ = 12.87km/h$

8 0

3 years ago

Find the slope of the line that passes through the points (-1, -2) and (-9, -2)

kodGreya [7K]

Slope = change in Y / change in x

Slope = (-2 - -2) / (-9 - -1)

Simplify:

Slope = 0/-8 = 0

Slope = 0

4 0

3 years ago