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3 years ago

8

True or false? If you took a true "if-then" statement, inserted a not in clause, and reversed the clauses, the new statement wou

ld also be true.

2 answers:

viva [34]3 years ago

8 0

Answer:

It's false

Step-by-step explanation:

Helga [31]3 years ago

4 0

If the statement is true, and you Rearrange it. It should still be true unless you added false facts into it.<span />

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Giselle works as a carpenter and as a blacksmith. She earns $20\$20 $20 dollar sign, 20 per hour as a carpenter and $25\$25 $25

malfutka [58]

C = # of carpenter hrs , b = # of blacksmith hrs

c + b = 30...c = 30 - b
20c + 25b = 690

20(30 - b) + 25b = 690
600 - 20b + 25b = 690
-20b + 25b = 690 - 600
5b = 90
b = 90/5
b = 18 <== hrs worked as blacksmith

c + b = 30
c + 18 = 30
c = 30 - 18
c = 12 <== hrs worked as carpenter

7 0

3 years ago

20(20-x)=300 what is x

neonofarm [45]

Answer:

x = 5

Step-by-step explanation:

20(20-x)=300

1. Divide both sides by 20.

20(20-x)/20 = 300/20

2. Simplify.

20-x=15

3. Subtract 20 from both sides.

20-x-29=15-20

4. Simplify.

-x=-5

5. Divide both sides by -1.

-x/-1 = -5/-1

6. Simplify.

x=5

5 0

2 years ago

Which situation is best? a. having $3,280.00 in a savings account earning a 4.25% APR and $1,320.00 due on a credit card with a

nexus9112 [7]

Answer:

consumer math / answer is A

Step-by-step explanation:

3 0

3 years ago

If ☆ x 2 = 0, then ☆ 8 will equal:<br> 0 + 6<br> O + 8<br> 0 x 2<br> O X4<br> 0 x 8

soldier1979 [14.2K]

The last one I think

3 0

2 years ago

Which equation has the solutions x=1+or-square root of 5?

stiv31 [10]

We will proceed to solve each case to determine the solution of the problem.

<u>case a)</u> $x^{2}+2x+4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}+2x=-4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}+2x+1=-4+1$

$x^{2}+2x+1=-3$

Rewrite as perfect squares

$(x+1)^{2}=-3$

$(x+1)=(+/-)\sqrt{-3}\\(x+1)=(+/-)\sqrt{3}i\\x=-1(+/-)\sqrt{3}i$

therefore

case a) is not the solution of the problem

<u>case b)</u> $x^{2}-2x+4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}-2x=-4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}-2x+1=-4+1$

$x^{2}-2x+1=-3$

Rewrite as perfect squares

$(x-1)^{2}=-3$

$(x-1)=(+/-)\sqrt{-3}\\(x-1)=(+/-)\sqrt{3}i\\x=1(+/-)\sqrt{3}i$

therefore

case b) is not the solution of the problem

<u>case c)</u> $x^{2}+2x-4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}+2x=4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}+2x+1=4+1$

$x^{2}+2x+1=5$

Rewrite as perfect squares

$(x+1)^{2}=5$

$(x+1)=(+/-)\sqrt{5}\\x=-1(+/-)\sqrt{5}$

therefore

case c) is not the solution of the problem

<u>case d)</u> $x^{2}-2x-4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}-2x=4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}-2x+1=4+1$

$x^{2}-2x+1=5$

Rewrite as perfect squares

$(x-1)^{2}=5$

$(x-1)=(+/-)\sqrt{5}\\x=1(+/-)\sqrt{5}$

therefore

case d) is the solution of the problem

therefore

<u>the answer is</u>

$x^{2}-2x-4=0$

5 0

3 years ago

Read 2 more answers