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3 years ago

Write the pair of fractions as a part of fractions with a common denominator 1/4 and 2/3

Mathematics

1 answer:

sladkih [1.3K]3 years ago

4 0

Answer:

$\frac{3}{12} and \frac{8}{12}$ are the fractions with common denominator

Step-by-step explanation:

fractions are 1/4 and 2/3

$\frac{1}{4} and \frac{2}{3}$

To get common denominator we take LCD

LCD of 4 and 3 is 12

To get 12 in the denominator we multiply first fraction by 3 and second fraction by 4

$\frac{1*3}{4*3} and \frac{2*4}{3*4}$

$\frac{3}{12} and \frac{8}{12}$

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If the measure of angle ABD and DBC are a linear pair, with a measure of angle ABD = (5x+29) and measure of angle DBC=(2x+32),wh

Nonamiya [84]

Answer:

x = 17

Step-by-step explanation:

5x + 29 + 2x + 32 = 180 (linear pair means that these angles add up to 180 cos they are in a line)

7x = 119

x = 17

8 0

3 years ago

3-(2x-5) +4<br><br> 4 +6(x+3)-(x-2)+7<br><br> -4x(x-2)+3-(6-2x)

Dvinal [7]

Answer:

1st solution is -2x+12

2nd solution is 5x+31

3rd one doesnt have a solution

Step-by-step explanation:

4 0

2 years ago

a triangular flower bed needs a thin metal border all the way around it. The sides are 7 feet, 6 feet, and 9 feet long.

kotegsom [21]

Excuse my awful triangle drawing skills, haha..
So essentially you need to find the perimeter which is done by adding the lengths of the sides.. this gives you 22ft.
Then for part b you need to convert 22ft to yards which gives you 7.3 yards.
Since you can only buy a whole number of yards, you'll actually need 8 yards.
So if the boarder is $8.75 per yard, and you need 8 yards, how much would it cost?
Hope this was helpful.

7 0

2 years ago

Initially a tank contains 10 liters of pure water. Brine of unknown (but constant) concentration of salt is flowing in at 1 lite

zhenek [66]

Answer:

Therefore the concentration of salt in the incoming brine is 1.73 g/L.

Step-by-step explanation:

Here the amount of incoming and outgoing of water are equal. Then the amount of water in the tank remain same = 10 liters.

Let the concentration of salt be a gram/L

Let the amount salt in the tank at any time t be Q(t).

$\frac{dQ}{dt} =\textrm {incoming rate - outgoing rate}$

Incoming rate = (a g/L)×(1 L/min)

=a g/min

The concentration of salt in the tank at any time t is = $\frac{Q(t)}{10}$ g/L

Outgoing rate = $(\frac{Q(t)}{10} g/L)(1 L/ min)$ $\frac{Q(t)}{10} g/min$

$\frac{dQ}{dt} = a- \frac{Q(t)}{10}$

$\Rightarrow \frac{dQ}{10a-Q(t)} =\frac{1}{10} dt$

Integrating both sides

$\Rightarrow \int \frac{dQ}{10a-Q(t)} =\int\frac{1}{10} dt$

$\Rightarrow -log|10a-Q(t)|=\frac{1}{10} t +c$ [ where c arbitrary constant]

Initial condition when t= 20 , Q(t)= 15 gram

$\Rightarrow -log|10a-15|=\frac{1}{10}\times 20 +c$

$\Rightarrow -log|10a-15|-2=c$

Therefore ,

$-log|10a-Q(t)|=\frac{1}{10} t -log|10a-15|-2$ .......(1)

In the starting time t=0 and Q(t)=0

Putting t=0 and Q(t)=0 in equation (1) we get

$- log|10a|= -log|10a-15| -2$

$\Rightarrow- log|10a|+log|10a-15|= -2$

$\Rightarrow log|\frac{10a-15}{10a}|= -2$

$\Rightarrow |\frac{10a-15}{10a}|=e ^{-2}$

$\Rightarrow 1-\frac{15}{10a} =e^{-2}$

$\Rightarrow \frac{15}{10a} =1-e^{-2}$

$\Rightarrow \frac{3}{2a} =1-e^{-2}$

$\Rightarrow2a= \frac{3}{1-e^{-2}}$

$\Rightarrow a = 1.73$

Therefore the concentration of salt in the incoming brine is 1.73 g/L

8 0

2 years ago

Determine if the following triangle is a right triangle or not using the Pythagorean Theorem Converse. Triangle with side length

seraphim [82]

Answer:

A right triangle is a triangle in which one of the angles is a 90∘ angle. The "square" at the vertex of the angle indicates that it is 90 degrees. A triangle can be determined to be a right triangle if the side lengths are known.

6 0

2 years ago