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Setler [38]
2 years ago
12

Which ratio is equivalent to the unit rate 30 miles 1 gallon ? How was the unit rate transformed into the equivalent ratio? A) 3

gallons 10 miles ; by dividing the numerator and denominator of the unit rate by 10. B) 90 miles 5 gallons ; by multiplying the numerator and denominator of the unit rate by 3. Eliminate C) 6 gallons 180 miles ; by multiplying the numerator and denominator of the unit rate by 2. D) 180 miles 6 gallons ; by multiplying the numerator and denominator of the unit rate by 6.
Mathematics
2 answers:
Nataly_w [17]2 years ago
5 0
It will be D by multiplying the numerator and the denominator by the unit rate of 6
solong [7]2 years ago
3 0
D)180 miles 6 gallons ; by multiplying the numerator and denominator of the unit rate by 6.
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3 years ago
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If 1/√a-√b=1/3 and 1/√a+√b=1/2, then find the difference of a and b.​
kow [346]

<u>ANSWER:</u>

If \frac{1}{\sqrt{a}-\sqrt{b}}=\frac{1}{3} and \frac{1}{\sqrt{a}+\sqrt{b}}=\frac{1}{2} then the difference of a and b is 6

<u>SOLUTION:</u>

Given, \frac{1}{\sqrt{a}-\sqrt{b}}=\frac{1}{3} →\sqrt{a}-\sqrt{b}=3 ----- (1)

And \frac{1}{\sqrt{a}+\sqrt{b}}=\frac{1}{2} → \sqrt{a}+\sqrt{b}=2 --- (2)

We have to find difference of a and b.

Now, add (1) and (2)

\sqrt{a}-\sqrt{b}=3

\sqrt{a}+\sqrt{b}=2

Adding above two equations, we get,

2 \sqrt{a}+0=2+3

\begin{array}{l}{2 \sqrt{a}=5} \\\\ {\sqrt{a}=\frac{5}{2}} \\\\ {a=\frac{25}{4}}\end{array}

substitute \sqrt{a} value in (2)

\begin{array}{l}{\frac{5}{2}+\sqrt{b}=2} \\\\ {\sqrt{b}=\frac{2}{\sin \frac{5}{2}}} \\\\ {\sqrt{b}=\frac{4-5}{2}} \\\\ {\sqrt{b}=\frac{-1}{2}} \\\\ {b=\frac{1}{4}}\end{array}

Now, difference of a and b is a – b = \frac{25}{4}-\frac{1}{4}=\frac{24}{4}=6

Hence, the difference of a and b is 6.

8 0
3 years ago
The average density of the material in intergalactic space is approximately 2.5 × 10–27 kg/m3. what is the volume of a lead samp
Lesechka [4]

Answer:

e. 1.8\times 10^{-6}m^3

Step-by-step explanation:

It is given that,

The density of intergalactic space material is 2.5 \times 10^{-27} kg per cubic meter.

And the volume of intergalactic space material is 8.0\times 10^{24} m^3

So the mass of that much intergalactic space material is,

m= \rho\times v, where 'm' is the mass, and 'v' is the volume.

Putting the values we get,

m=2.5 \times 10^{-27}\times 8.0 \times 10^{24}

m=20\times 10^{(-27+24)}=20\times 10^{-3} kg

It is also given that the mass of lead is the same as the mass of the intergalactic space material. Therefore, the mass of lead is 20 \times 10^{-3}=2 \times 10^{-2}kg

So the volume of lead is,

v=\frac{m}{\rho} =\frac{2 \times 10^{-2}}{11300} = 0.00000177 m^3

v=1.77 \times 10^{-6}=1.8\times 10^{-6}m^3

So the volume of lead is v=1.8\times 10^{-6}m^3.

3 0
3 years ago
What is the solution of this equation?<br> -1y = -4
tigry1 [53]

Answer:

y = 4

Step-by-step explanation:

-1y = -4

Divide both sides by -1 to get y by itself.

-1/-1 = -4/-1

y = 4

It's positive because there are two negatives, which mean they cancel out.

8 0
3 years ago
Write the sequence of transformations that changes figure ABCD to figure A'B'C'D'.
creativ13 [48]

Answer:

(x-2)(y+1)

Step-by-step explanation:

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2 years ago
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