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2 years ago

6

i promise i will add you to brainliest if anyone get the answer correct promise and show steps....please help I am taking a test

And i know you could do it

1 answer:

Alexandra [31]2 years ago

7 0

Answer:

5.2 liters

Step-by-step explanation:

3.9x3=11.7

3.9+11.7=15.6

15.6/3=5.2

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Quickly solve (x+7)2=121, for x. Graph the solution on a number line.

asambeis [7]

Answer:

The value of the x is 53.5 or x=53.5

Step-by-step explanation:

Given data in the question:-

$(x+7)2=121$

We have to find the value of x.

Solution:-

$(x+7)2=121\\=>2x+14=121\\=>2x=121-14\\=>2x=107\\=>x=107/2\\=>x=53.5\\$

Hence the answer is 53.5

3 0

2 years ago

Determine whether the given value is a solution of the equation: x + 15 = 10; x=5 *

kobusy [5.1K]

Answer:

It is not correct, x = -5

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

Help me please. I am having troubles with my math homework.

jeyben [28]

Answer: (3) f(8) = g(8)

<u>Step-by-step explanation:</u>

Let's compare the values of f(x) and g(x) when x = 0, 2, 8, and 4

<u> f(x) </u> <u> g(x) </u> <u>Comparison</u>

f(x) = 2x - 3 $g(x)=\dfrac{3}{2}x+1$

f(0) = 2(0) - 3 $g(0)=\dfrac{3}{2}(0)+1$

= -3 = 1 f(0) < g(0)

f(2) = 2(2) - 3 $g(2)=\dfrac{3}{2}(2)+1$

= 1 = 4 f(2) < g(2)

f(8) = 2(8) - 3 $g(8)=\dfrac{3}{2}(8)+1$

= 13 = 13 f(8) = g(8)

f(4) = 2(4) - 3 $g(4)=\dfrac{3}{2}(4)+1$

= 5 = 7 f(4) < g(4)

The only statement provided that is true is f(8) = g(8)

3 0

2 years ago

What dose this equal 2(6x-4)=3(6x+2)

AveGali [126]

Answer:

-7/3

Step-by-step explanation:

2(6−4)=3(6+2)

2(6x-4)=3(6x+2)

Solve

1

Distribute

2(6−4)=3(6+2)

{\color{#c92786}{2(6x-4)}}=3(6x+2)

12−8=3(6+2)

{\color{#c92786}{12x-8}}=3(6x+2)

2

Distribute

12−8=3(6+2)

12x-8={\color{#c92786}{3(6x+2)}}

12−8=18+6

12x-8={\color{#c92786}{18x+6}}

3

Add

8

8

to both sides of the equation

12−8=18+6

12x-8=18x+6

12−8+8=18+6+8

12x-8+{\color{#c92786}{8}}=18x+6+{\color{#c92786}{8}}

5 more steps

Solution

=−7/3

7 0

2 years ago

Read 2 more answers

Suppose that five ones and four zeros are arranged around a circle. Between any two equal bits you insert a 0 and between any tw

PolarNik [594]

Answer:

Using <u>backward reasoning</u> we want to show that <em>"We can never get nine 0's"</em>.

Step-by-step explanation:

Basically in order to create nine 0's, the previous step had to have all 0's or all 1's. There is no other way possible, because between any two equal bits you insert a 0.

If we consider two cases for the second-to-last step:

<u>There were 9 </u><u>0's</u><u>:</u>

We obtain nine 0's if all bits in the previous step were the same, thus all bit were 0's or all bits were 1's. If the previous step contained all 0's, then we have the same case as the current iteration step. Since initially the circle did not contain only 0's, the circle had to contain something else than only 0's at some point and thus there exists a point where the circle contained only 1's.

<u>There were 9 </u><u>1's</u><u>:</u>

A circle contains only 1's, if every pair of the consecutive nine digits is different. However this is impossible, because there are five 1's and four 0's (we have an odd number of bits!), thus if the 1's and 0's alternate, then we obtain that 1's that will be next to each other (which would result in a 1 in the next step). Thus, we obtained a contradiction and thus assumption that the circle contains nine 0's after iteratins the procedure is false. This then means that you can never get nine 0's.

To summarize, in order to create nine 0's, the previous step had to have all 0's or al 1's. As we didn't start the arrange with all 0's, the only way is having all 1's, but having all 1's will not be possible in our case since we have an odd number of bits.

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5 0

3 years ago