All categories

3 years ago

What is the solution to 2x+5>3x+1

Mathematics

2 answers:

frez [133]3 years ago

7 0

   2x + 5 > 3x + 1
<u> - 2x     - 2x </u>
   5 > x + 1
   <u>- 1 - 1</u>
   4 > x
   x < 4

Colt1911 [192]3 years ago

4 0

The solution is 4>x Because you subtract 1 from both sides and you subtract 2x from both sides and you get x<4 or 4>x

You might be interested in

4 multiple ( 2+3) square 2

shtirl [24]

Answer:

100

Step-by-step explanation:

4 x (2 + 3) $^{2}$

for this equation u have to use PEMDAS so u have to solve the parenthesis first which is 5 then exponets which is $5^{2}$ which is 25 and 24 x 4 gives you 100

hope this helps, if im wrong sorry

7 0

2 years ago

Read 2 more answers

What is the common ratio between successive terms in the sequence?

amid [387]

Answer:

Common Ratio=1/3

Step-by-step explanation:

4 0

2 years ago

I need major help with 3 and 4 PLEASE

Westkost [7]

Answer:

I answered 3 (your recent question)

The answer to 4 is C

Step-by-step explanation:

MRQL is the plane on the right hand side.

Lines parallel to that are: JK, NP, JN, and KP.

Thus the correct answer is C.

5 0

2 years ago

The math team has 11 boys, 10 girls, and 1 coach. What is the ratio of coach to students? Group of answer choices 21/1 11/1 1/11

laila [671]

Answer:

1/21

Step-by-step explanation:

11 boys + 10 girls = 21 students

We want the ratio of coaches to students

1 coach: 21 students

1/21

8 0

3 years ago

Read 2 more answers

A triangle has a height that is increasing at a rate of 2 cm/sec and its area is increasing at a rate of 4 cm2/sec. Find the rat

lesya [120]

Answer:

The value of rate of which the base is changing $\frac{dB}{dt}$ = - 3 $\frac{cm}{s}$

Step-by-step explanation:

Area = 20 $cm^{2}$

height = 4 cm

$\frac{dH}{dt} = 2 \frac{cm}{sec}$

$\frac{dA}{dt} =$ $2 \ \frac{cm^{2} }{sec}$

we know that area of the triangle is given by

$A= \frac{1}{2} B H$

$B = \frac{2A}{H}$

$B = \frac{2 (20)}{4}$

B = 10 cm

Rate of change of area is given by

$\frac{dA}{dT} = \frac{1}{2} [B\frac{dH}{dt} + H \frac{dB}{dt} ]$

4 = 0.5 [10 × 2 + 4 $\frac{dB}{dt}$ ]

$\frac{dB}{dt}$ = - 3 $\frac{cm}{s}$

This is the value of rate of which the base is changing.

7 0

2 years ago