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3 years ago

Determine which ratio forms a proportion with 7/2

Mathematics

2 answers:

SCORPION-xisa [38]3 years ago

7 0

You should multiply it by the same number such as 14/2 to 7/2. Otherwise, it simply does not work.

gizmo_the_mogwai [7]3 years ago

5 0

For determining proportional ratios, you have to multiply both the denominator and the numerator by the same number

Eg. 14/4 is proportional to 7/2 while 22/6 is not proportional to 7/2.

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Select the correct inequality for the graph given

Kaylis [27]

Answer:

y<= 3x-1

Step-by-step explanation:

When it comes about inequalities, keep in mind the following

the < sign means that the regions is under whatever is to other side, and the > means that the region is above whatever is on the other side of the sign

So, we see for the graphic that there is a region below a line, this suggests us that there is a < sign implied

The other thing is that there is a convention in math, is the line is graphed using a dashed line, it meas that it is not part of the region, if the line is graphed using a solid line (a continued line) then the line is inside the region, hence there is a = sign implied.

So the answer is y<= 3x-1

4 0

3 years ago

What is the value of -7/8 divided by -1 2/5

Zigmanuir [339]

(-7/8) / (-1 2/5) = (-7/8) / (-7/5) = (-7/8) * (-5/7) = 5/8

4 0

3 years ago

I Need Assistance With This:

Tamiku [17]

Answer:

22.5 in^2

Step-by-step explanation:

h = 3 in

base 1 = 6 in

base 2 = 9 in

So plug in

A = 1/2 (3) (6 + 9)

A = 3/2 (15)

A = 22.5 in^2

5 0

3 years ago

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

4 years ago

Complete the statement by typing the letter of the angle.<br> m∠ =54º

tekilochka [14]

Answer:

x = 16

∠A = 39

∠B = 54

∠C = 87

Step-by-step explanation:

Make a equation of all of the angles added together is equal to 180 since the angles of a triangle added together is 180

(4x-10)+(2x+7)+(7x-25)=180

combine like terms

4x+2x+7x-10+7-25 = 180

13x-28=180

solve for x

13x=208

x = 16

plug in x to find the measure of the angle

∠A = 2(16) + 7 = 39

∠B = 4(16) - 10 = 54

∠C = 7(16) - 25 = 87

Add them together to double check your answer

39 + 54 + 87 = 180

hope this helps

3 0

3 years ago

Read 2 more answers