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3 years ago

The solution to 2x+8>10 is x<1 true or false

Mathematics

2 answers:

Travka [436]3 years ago

8 0

Answer:

false.

Step-by-step explanation:

Given : 2x+8>10.

To find : the solution to 2x+8>10 is x<1 true or false.

Solution : We have given that

2x+8>10.

On subtracting both sides by 8

2x > 10 -8.

2x > 2

On dividing both sides by 2

x > 1

So, given statement false .

Therefore, false.

Morgarella [4.7K]3 years ago

5 0

Answer:

false

Step-by-step explanation:

correct me if this is wrong, but it's actually x>1

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The fastest speed a table tennis ball has been hit about 13.07 times as fast as the speed for the fastest swimming. What is the

Effectus [21]

v = speed of fastest swimming = 8.6 kilometer/hour

V = fastest speed of a tennis ball = ?

It is given in the question that

fastest speed of a tennis ball = (13.07) speed of fastest swimming

V = (13.07) v

inserting the values

V = (13.07) (8.6 kilometer/hour)

V = (13.07 x (8.6) kilometer/hour)

V = 112.4 kilometer/hour

6 0

2 years ago

What does it equal to?

Marina CMI [18]

Answer:

Step-by-step explanation:

|x| + |x-2| =

Let x= 1

|1| + |1-2| =

|1| + |-1| =

The absolute value means take the non negative value

1 + 1 = 2

8 0

3 years ago

Read 2 more answers

Ax+by=(a-b) , bx - ay =(a+b) simultaneous linear equations using cross multiplication

Molodets [167]

Answer:

$x=1\,,\,y=-1$

Step-by-step explanation:

Given: $ax+by=a-b\,,\,bx-ay=a+b$

To solve: the given linear equations

Solution:

Consider the equations:

$A_1x+B_1y+C_1=0\\A_2x+B_2y+C_2=0$

By method of cross multiplication:

$\frac{x}{B_1C_2-B_2C_1}=\frac{y}{C_1A_2-C_2A_1}=\frac{1}{A_1B_2-A_2B_1}$

For equations: $ax+by=a-b\,,\,bx-ay=a+b$

$ax+by-(a-b)=0\\bx-ay-(a+b)=0$

Take $A_1=a\,,\,B_1=b\,,\,C_1=-(a-b)\,,\,A_2=b\,,\,B_2=-a\,,\,C_2=-(a+b)$

So,

$\frac{x}{-b(a+b)-a(a-b)}=\frac{y}{-b(a-b)+a(a+b)}=\frac{1}{-a^2-b^2}\\\frac{x}{-ab-b^2-a^2+ab}=\frac{y}{-ab+b^2+a^2+ab}=\frac{1}{-(a^2+b^2)}\\\frac{x}{-(a^2+b^2)}=\frac{y}{a^2+b^2}=\frac{1}{-(a^2+b^2)}\\\frac{x}{-(a^2+b^2)}=\frac{1}{-(a^2+b^2)}\,,\,\frac{y}{a^2+b^2}=\frac{1}{-(a^2+b^2)}\\x=1\,,\,y=-1$