All categories

2 years ago

How many solutions are there to the following equation?

Mathematics

2 answers:

Hoochie [10]2 years ago

6 0

Answer:

D) infinitely many solutions

Step-by-step explanation:

5 ( x-3 ) - 3x = 8x - 15 - 6x

5x - 3x - 15 = 8x - 6x - 15

2x - 15 = 2x - 15

2x = 2x

Since, equations of both lines are same. Therefore, there are infinitely many solutions.

romanna [79]2 years ago

3 0

Answer:

Step-by-step explanation:

Given

5(x - 3) - 3x = 8x - 15 - 6x

5x - 15 - 3x = 2x - 15

2x - 15 = 2x - 15

Since both sides are the same then any real value of x will be a solution.

That is there are infinitely many solutions

You might be interested in

How many terms in 9a+11p

Naddik [55]

Answer:

2 terms

Step-by-step explanation:

Terms are nothing but the mixtures of co-effecients and their variables..

Here, 9a is one term while 11p is the other

7 0

2 years ago

Read 2 more answers

B-3±6-2b i need it know pls help me

aleksklad [387]

Answer : b=3

Hope this helps have a great day! :)

3 0

3 years ago

Three different divers kept track of the number of treasure boxes they've found this year.

steposvetlana [31]

Choice a i'm pretty sure !

7 0

2 years ago

Read 2 more answers

How far away is a star with a parallax of 2 arc sec in parsecs? In AU?

Vesna [10]

Answer:

ear

Step-by-step explanation:

8 0

3 years ago

Triangle ABC has AB=5, BC=7, and AC=9. D is on AC with BD=5. Find the length of DC

e-lub [12.9K]

Answer:

$DC=\frac{8}{3}\ units$

Step-by-step explanation:

The picture of the question in the attached figure

we know that

The triangle ABD is an isosceles triangle

because

AB=BD

The segment BM is a perpendicular bisector segment AD

In the right triangle ABM

Applying the Pythagorean Theorem

$BM^2=AB^2-AM^2$

we have

$AB=5\ units\\AM=x\ units$

substitute

$BM^2=5^2-x^2$

$BM^2=25-x^2$ -----> equation A

In the right triangle BMC

Applying the Pythagorean Theorem

$BM^2=BC^2-MC^2$

we have

$BC=7\ units\\MC=AC-AM=(9-x)\ units$

substitute

$BM^2=7^2-(9-x)^2$

$BM^2=49-(81-18x+x^2)$

$BM^2=49-81+18x-x^2$

$BM^2=-x^2+18x-32$ ----> equation B

equate equation A and equation B

$-x^2+18x-32=25-x^2$

solve for x

$18x=25+32\\18x=57\\\\x=\frac{57}{18}$

Simplify

$x=\frac{19}{6}$

Find the length of DC

$DC=AC-2x$

substitute the given values

$DC=9-2(\frac{19}{6})$

$DC=9-\frac{19}{3}\\\\DC=\frac{8}{3}\ units$

8 0

3 years ago