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3 years ago

What is the solution to the equation?

2 answers:

7 0

B.
X=5

7(5+9)=7•x+7•9
(7•5)+(7•9)=7x+63
35+63=7x+63

Subtract 63 from both sides and you will be left with
35=7x
Divide each side by 7 (you pretty much just divide 35 by 7)

And x=5

Lesechka [4]3 years ago

7 0

To get the answer we have to solve eq.
7(5+9)=7x+7*9
7*5+7*9=7x+63
35+63=7x+63 /-63
35=7x /:7
5=x
The correct answer is B.

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What is the area of this composite figure?

Gelneren [198K]

Answer:

249 centimeters squared

Step-by-step Explanation:

The area of the rectangle: l x w, where l is the length and w is the width.

= 15.5 x 18

= 279

Area of the smaller rectangle: l x w, where l is the length and w is the width.

= 4 x 7.5

= 30

We do not need this 30.

Area of the shaded region:

= 279 - 30

= 249 cm²

5 0

2 years ago

Which system of equations below has infinitely many solutions? y = –3x 4 and y = –3x – 4 y = –3x 4 and 3y = –9x 12 y = –3x 4 and

Flura [38]

the equations y = –3x + 4 and 3y = –9x + 12 have infinitely many solutions. option B is correct.

<h3>What is the linear system?</h3>

It is a system of an equation in which the highest power of the variable is always 1. A one-dimension figure that has no width. It is a combination of infinite points side by side.

Condition for the parallel lines.

L1, ax + bx + c = 0

L2, dx + ey + f = 0

If $\rm \dfrac{a}{d} = \dfrac{b}{e} = \dfrac{c}{f}$ then lines have infinitely many solutions.

<h3>Which system of equations below has infinitely many solutions?</h3>

y = –3x + 4 and 3y = –9x + 12

On comparing we have

a = -3 , b = 1, and c = 4

d = -9 , e = 3, and f = 12

Then their ratio will be

$\rm \dfrac{1}{3} = \dfrac{-3}{-9} = \dfrac{4}{12}\\\\\rm \dfrac{1}{3} = \dfrac{1}{3} = \dfrac{1}{3}$

Hence y = –3x + 4 and 3y = –9x + 12 have infinitely many solutions.

Thus the option B is correct.

More about the linear system link is given below.

brainly.com/question/20379472

7 0

2 years ago

If I had 22 apples and someone stole 5. How many would that person have?

Sunny_sXe [5.5K]

That person, the someone, would have the 5 that they stole

hope this helps ;)

3 0

3 years ago

Read 2 more answers

The half-life of a radioactive element is 130 days, but your sample will not be useful to you after 80% of the radioactive nu

gtnhenbr [62]

Answer:

We can use the sample about 42 days.

Step-by-step explanation:

Decay Equation:

$\frac{dN}{dt}\propto -N$

$\Rightarrow \frac{dN}{dt} =-\lambda N$

$\Rightarrow \frac{dN}{N} =-\lambda dt$

Integrating both sides

$\int \frac{dN}{N} =\int\lambda dt$

$\Rightarrow ln|N|=-\lambda t+c$

When t=0, N= $N_0$ = initial amount

$\Rightarrow ln|N_0|=-\lambda .0+c$

$\Rightarrow c= ln|N_0|$

$\therefore ln|N|=-\lambda t+ln|N_0|$

$\Rightarrow ln|N|-ln|N_0|=-\lambda t$

$\Rightarrow ln|\frac{N}{N_0}|=-\lambda t$ .......(1)

$\frac{N}{N_0}=e^{-\lambda t}$ .........(2)

Logarithm:

$ln|\frac mn|= ln|m|-ln|n|$
$ln|ab|=ln|a|+ln|b|$

$ln|e^a|=a$

$ln|a|=b \Rightarrow a=e^b$
$ln|1|=0$

130 days is the half-life of the given radioactive element.

For half life,

$N=\frac12 N_0$ , $t=t_\frac12=130$ days.

we plug all values in equation (1)

$ln|\frac{\frac12N_0}{N_0}|=-\lambda \times 130$

$\rightarrow ln|\frac{\frac12}{1}|=-\lambda \times 130$

$\rightarrow ln|1|-ln|2|-ln|1|=-\lambda \times 130$

$\rightarrow -ln|2|=-\lambda \times 130$

$\rightarrow \lambda= \frac{-ln|2|}{-130}$

$\rightarrow \lambda= \frac{ln|2|}{130}$

We need to find the time when the sample remains 80% of its original.

$N=\frac{80}{100}N_0$

$\therefore ln|{\frac{\frac {80}{100}N_0}{N_0}|=-\frac{ln2}{130}t$

$\Rightarrow ln|{{\frac {80}{100}|=-\frac{ln2}{130}t$

$\Rightarrow ln|{{ {80}|-ln|{100}|=-\frac{ln2}{130}t$

$\Rightarrow t=\frac{ln|80|-ln|100|}{-\frac{ln|2|}{130}}$

$\Rightarrow t=\frac{(ln|80|-ln|100|)\times 130}{-{ln|2|}}$

$\Rightarrow t\approx 42$

We can use the sample about 42 days.

7 0

3 years ago

-1 1/4 + 1/2 <br> A)-1 3/4 <br> B)-1 1/2 <br> C)-3/4 <br> D)-1/2

bearhunter [10]

Answer:

<em>Well, Your best answer will be is </em><em>B. -1 1/2. </em>

<em>Hope That Helps!</em>

$^{Itsbrazts}$

8 0

3 years ago