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3 years ago

12

Please help-help-help

1 answer:

Mrac [35]3 years ago

8 0

Answer:

both I think hope u get it right

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Ent<br> =<br> Solve for x.<br> 4(x + 1) < 4x + 3

Igoryamba

Answer:

x=allrealsolutions

Step-by-step explanation:

5 0

3 years ago

Kelly went to a hockey game with her family. They bought 4 bags of cotton candy and 2 souvenir cups and spent $36. Marissa's fam

erik [133]

4b+2c=36, solve for b, subtract 2c from both sides

4b=36-2c divide both sides by 4

b=9-0.5c

Then we are told that:

7b+3c=59, using b=9-0.5c we get:

7(9-0.5c)+3c=59 perform indicated multiplication on left side

63-3.5c+3c=59 combine like terms on left side

63-0.5c=59 subtract 63 from both sides

-0.5c=-4 divide both sides by 0.5

c=8

So each souvenir cup costs $8.

(and each bag of cotton candy is $5)

3 0

3 years ago

Triss [41]

Answer:

42390 ft³

Step-by-step explanation:

Since the tanks are congruent, they both have same volume.

If you rejoin them, it will form a cylinder with the radius of 15 ft and height of 120 ft.

<u>The volume of the cylinder:</u>

V = πr²h = 3.14*15²*120 = 84780 ft³

<u>The volume of each tank is half the volume of the cylinder:</u>

V(tank) = 84780 / 2 = 42390 ft³

5 0

3 years ago

Read 2 more answers

Simplify: 6 x (-5)+(-18)

faust18 [17]

Answer:

-138

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Can someone help me out with some pesky math quiz :) I put a pic

Sindrei [870]

1. Simplifying ∛54 gives 3∛2

2. 27^1/3 * (27)^3 is equal to 27

3. The expression (2j4k)^4/5 can be written as $\sqrt[4]{32j^{20}k^{5} }^{5}$ making the first option the correct option

4. simplifying $\sqrt[4]{16x^{4} y^{12} }$ will result to 2x∛y

5. simplifying $\sqrt[5]{2940x^{13} y^{7} }$ will result to $14x^{6} y^{3} \sqrt{15xy}$ making the last option the correct option

<h3>How to simplify $\sqrt[5]{2940x^{13} y^{7} }$ </h3>

<u>given data</u>

$\sqrt[5]{2940x^{13} y^{7} }$

= ( 2940 * x^13 * y^7 )^1/5

$= ( 2^2 * 5 * 3 * 7^2 * x^{12} * x * y^6 * y )^{1/5}$

$= ( 2^2 * 5 * 3 * 7^2 * x^{12} * x * y^6 * y )^{1/5*5/2}$

$= ( 2^2 * 5 * 3 * 7^2 * x^{12} * x * y^6 * y )^{5/5*1/2}$

multiplying the values inside the parenthesis by ^1/2

$= ( 2 * 5^{1/2} * 3^{1/2} * 7 * x^{6} * x^{1/2} * y^3 * y^{1/2} )$

$= ( 2 * 5^{1/2} * 3^{1/2} * 7 * x^{6} * x^{1/2} * y^3 * y^{1/2} )$

$= ( 14* 5^{1/2} * 3^{1/2} * x^{6} * x^{1/2} * y^3 * y^{1/2} )$

$= ( 14 * x^6 * y^3 ) ( 5^{1/2} * 3^{1/2} * x^{1/2} * y^{1/2} )$

$= ( 14x^6y^3 ) ( 15^{1/2} * x^{1/2} * y^{1/2} )$

$= ( 14x^6y^3 ) ( 15xy)^{1/2}$

$= ( 14x^6y^3 ) \sqrt{( 15xy)}$

we can therefore say that the last option is the answer

Read more on equations here: brainly.com/question/22688504

#SPJ1

3 0

1 year ago