All categories

3 years ago

How do you do these problems

Mathematics

1 answer:

avanturin [10]3 years ago

7 0

$\cfrac{1}{3} = \cfrac{2}{6} = \cfrac{3}{9} = \cfrac{6}{18}$

$\cfrac{4}{20} = \cfrac{3}{15} = \cfrac{12}{60} = \cfrac{100}{500}$

You might be interested in

What would be identified as a constant(s) in: 7h + 3

Sveta_85 [38]

Answer:

Number 3 would be considered a constant.

Step-by-step explanation:

7h and an h at the end, so it isn't a constant number.

8 0

3 years ago

Y is inversely proportional to x.<br> When x=9, y = 8.<br> Find y when x= 6.

Pani-rosa [81]

Answer:

it should be 5

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

Please help with this math problem

zimovet [89]

$x=2+½*\sqrt{10}$

$x=2-½*\sqrt{10}$

Answer:

Solution given:

y=2x²-5x+6.....[1]

y=3x+3.........….[2]

solving equation 1&2

2x²-5x+6=3x+3

2x²-5x-3x+6-3=0

2x²-8x+3=0

comparing above equation with ax²+bx+c=0 we get

a=2

b=-8

c=3

by using quadratic equation

$x=\frac{+8±\sqrt{(-8)²-4*2*3}}{2*2}$

$x=\frac{+8±2\sqrt{10}}{4}$

taking positive

$x=\frac{+8+2\sqrt{10}}{4}$

$x=2+½*\sqrt{10}$

taking negative

$x=2-½*\sqrt{10}$

3 0

3 years ago

Please answer this question fast

Sphinxa [80]

Answer:

I think it is 6(5/6)

6 0

3 years ago

At the moment a hot cake is put in a cooler, the difference between the cake's and the cooler's temperature is 50° Celsius. This

dybincka [34]

Answer:

$D(t)=50(0.8)^t$

Step-by-step explanation:

We are given that

Initially the difference between the cake's and the cooler's temperature ,a=50 degree Celsius

$r=\frac{1}{5}/min$

We have to find the function that gives the temperature difference in degrees Celsius D(t).

We know that

$D(t)=a(1-r)^t$

Substitute the values

$D(t)=50(1-\frac{1}{5})^t=50(1-0.2)^t$

$D(t)=50(0.8)^t$

This is required function that gives the temperature difference in degrees Celsius.

6 0

3 years ago

Read 2 more answers