All categories

2 years ago

How do I find the length of GJ?

Mathematics

1 answer:

wlad13 [49]2 years ago

5 0

6(X+6)= 12(5)
6x+36=60
6x=24
X=4
Now put 4 in X of GJ
GJ=10

You might be interested in

What is the domain of the function y=2√x-6?<br> 88AX<∞<br> O 0≤x<∞<br> 3≤x<∞<br> 6≤x<∞

Travka [436]

Answer:

If the x-6 is all under the sqrt, the last answer is right:

6<= x <= infinity

If only x is under the sqrt and - 6 is NOT under the sqrt, then:

0 <= x <= infinity is the correct answer.

Step-by-step explanation:

You can look at the equation and or graph to determine the domain of a function. The domain is all the x's that may go into the equation or all the x's on the graph of the function. Inside of a squareroot symbol the number inside there must be positive. While considering the equation, only x's that keep that inside positive can be used.

For sqrt(x-6), x has to be 6 or bigger.

For sqrt(x) - 6, x has to be 0 or bigger.

6 0

1 year ago

How do i solve this?

Black_prince [1.1K]

No beacuse if you multiply 10 by 3 tents you get 1,000 your supposed to multiply 3 ones

8 0

3 years ago

a boat travels upstream on the allegheny river for 2 hours. the return trip only takes 1.7 hours because the boat travels 2.5 mi

Gre4nikov [31]

To solve this we are going to use the speed equation: $S= \frac{d}{t}$
where
$S$ is speed
$d$ is distance
$t$ time

We know from our problem that the upstream trip takes 2 hours, so $t_{u}=2$ . We also know that the downstream trip takes 1.7 hours, so $t_{d}=1.7$ . Notice that the distance of both trips is the same, so we are going to use $d$ to represent that distance.
Now, lets use our equation to relate the quantities:

For the upstream trip:
$S_{u}= \frac{d}{t_{u} }$
$S_{u}= \frac{d}{2}$ equation (1)

For the downstream trip:
$S_{d}= \frac{d}{t_{d} }$
$S_{d}= \frac{d}{1.7}$ equation (2)

We know that the boat travels 2.5 miles per hour faster downstream, so the speed of the boat upstream will be the speed of the boat downstream minus 2.5 miles per hour:
$S_{u}=S_{d}-2.5$ equation (3)

Replacing (3) in (1):
$S_{u}= \frac{d}{2}$
$S_{d}-2.5= \frac{d}{2}$ equation (4)

Solving for $d$ in equation (2):
$S_{d}= \frac{d}{1.7}$
$d=1.7S_{d}$ equation (5)

Replacing (5) in (4):
$S_{d}-2.5= \frac{d}{2}$
$S_{d}-2.5= \frac{1.7S_{d}}{2}$
$2(S_{d}-2.5)=1.7S_{d}$
$2S_{d}-5=1.7S_{d}$
$0.3S_{d}=5$
$S__{d}= \frac{5}{0.3}$
$S_{d}= \frac{50}{3}$ equation (6)

Replacing (6) in (5)
$d=1.7S_{d}$
$d=1.7( \frac{50}{3} )$
$d= \frac{85}{3}$ miles

We can conclude that the boat travel $\frac{85}{3}$ , which is approximately 28.3 miles, in one way.

5 0

3 years ago

Adding (4-7i)+(-11+9i)

Fofino [41]

your answer is -7+16i, because 4-11=-7 and 7i+9i is 16i

5 0

2 years ago

Please help. Ignore The answer I have selected. I'm not sure if it's correct

Karo-lina-s [1.5K]

Yes, I believe that is correct.

5 0

3 years ago