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3 years ago

What is the solution to the open sentence 18 = c + 6?

Mathematics

2 answers:

True [87]3 years ago

8 0

Subtract 6 from both side wich will leaveyou with a answer of 12=c

sveticcg [70]3 years ago

4 0

This is very simple, all you have to do is subtract 18 and 6 and that would be 12, and now we have 12 as a POSSIBLE answer, lets check our work to be 100% correct so now try to add 12 and 6 and we get 18, so yay it made the equation true and since its true its right so that's the answer.

Answer: c = 12 so 18 = 12 + 6 is true.

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Frank wrote these ratios to describe the number of votes each candidate received compared to the total. Which one is wrong, and

strojnjashka [21]

Answer:The right answer is (B)

Step-by-step explanation:

Your welcome

3 0

4 years ago

What is the slope of у>- 2х + 3

charle [14.2K]

Answer:

-2x

Step-by-step explanation:

Graph

y > −2x + 3

Use the slope-intercept form to find the slope and y-intercept.

The slope-intercept form is y = mx + b, where m is the slope and b is the y-intercept.

y = mx + b

Find the values of m and b using the form y = mx + b. m = −2

b = 3

The slope of the line is the value of m, and the y-intercept is the value of b.

Slope: −2

intercept: (0, 3)

Graph a dashed line, then shade the area above the boundary line since y is greater than

−2x + 3.

y > −2x + 3

8 0

3 years ago

SOMEONE PLEASE HELP ME ASAP PLEASE!!!!

ehidna [41]

Answer:

Therefore,

$a_{21}=2$

Step-by-step explanation:

Given:

$A=\left[\begin{array}{ccc}3&6&9\\2&4&8\\\end{array}\right]$

To Find:

a₂₁ = ?

Solution:

Let,

$A=\left[\begin{array}{ccc}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\\end{array}\right]$

We require ' a₂₁ ' i.e Second Row First Column Element

So on Comparing we get

∴ $a_{21}=2$

Therefore,

$a_{21}=2$

5 0

3 years ago

The radioactive isotope radium has a half-life of 1,600 years. The mass of a 100-gram sample of radium will be______ grams after

Olenka [21]

To solve this we are going to use the half life equation $N(t)=N_{0} e^{( \frac{-0.693t}{t _{1/2} }) }$
Where:
$N_{0}$ is the initial sample
$t$ is the time in years
$t_{1/2}$ is the half life of the substance
$N(t)$ is the remainder quantity after $t$ years

From the problem we know that:
$N_{0} =100$
$t=200$
$t_{1/2} =1600$

Lets replace those values in our equation to find $N(t)$ :
$N(200) =100e^{( \frac{(-0.693)(200)}{1600}) }$
$N(200)=100e^{( \frac{-138.6}{1600} )}$
$N(200)=100e^{-0.086625}$
$N(200)=91.7$

We can conclude that after 1600 years of radioactive decay, the mass of the 100-gram sample will be 91.7 grams.

8 0

3 years ago

What is the gcf of 28 and 60

Vikki [24]

Answer:

Step-by-step explanation:

We need to find the factors of each number

28 -

1x28

2x14

4x7

60 -

1x60

2x30

3x20

4x15

5x12

6x10

then we need to find the largest number they have in common

in this case it is 4

3 0

4 years ago