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Delicious77 [7]

3 years ago

14

Lauren is running in a race to raise money for her favorite charity the total distance of the race is 12.5 miles so far she has

run 1/10 of the race how far has lauren run

1 answer:

solmaris [256]3 years ago

6 0

To get the answer I divided 12.5 by 10. The answer is 1.25.

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A) Find the value of 3a + 4b when a = 17 and b = 9.

Ivan

Answer:

A. 87

B. i dont uderstand velocily sorry

Step-by-step explanation:

3*a=3*17=51

4*b=4*9=36

51+36=87

4 0

3 years ago

Read 2 more answers

Glenn bought 3 pounds of tomatoes. He used § of them to make sauce.

Darya [45]

Answer:i dont kno what number u put

Step-by-step explanation:

8 0

3 years ago

Please help lot of points at stake- and will give brainly.

Veronika [31]

Answer:

a)1400 + 300x ≤ 5000

b) x ≤ 12 days

Step-by-step explanation:

a) Since the grocery store owner wants to stay within budget then, this means he would be spending at most $5,000

The word at most is represented by the inequality sign ≤ = Less than or equal to

Let x = Number of days

Hence,

Our inequality equation is

$1400 + 300× x ≤ $5000

1400 + 300x ≤ 5000

b) Solving for x

1400 + 300x ≤ 5000

300x ≤ 5000 - 1400

300x ≤ 3600

x ≤ 3600/300

x ≤ 12 days

3 0

3 years ago

Suppose you have a light bulb that emits 50 watts of visible light. (Note: This is not the case for a standard 50-watt light bul

natali 33 [55]

Answer:

0.049168726 light-years

Step-by-step explanation:

The apparent brightness of a star is

$\bf B=\displaystyle\frac{L}{4\pi d^2}$

where

<em>L = luminosity of the star (related to the Sun) </em>

<em>d = distance in ly (light-years) </em>

The luminosity of Alpha Centauri A is 1.519 and its distance is 4.37 ly.

Hence the apparent brightness of Alpha Centauri A is

$\bf B=\displaystyle\frac{1.519}{4\pi (4.37)^2}=0.006329728$

According to the inverse square law for light intensity

$\bf \displaystyle\frac{I_1}{I_2}=\displaystyle\frac{d_2^2}{d_1^2}$

where

$\bf I_1=$ light intensity at distance $\bf d_1$

$\bf I_2=$ light intensity at distance $\bf d_2$

Let $\bf d_2$ be the distance we would have to place the 50-watt bulb, then replacing in the formula

$\bf \displaystyle\frac{0.006329728}{50}=\displaystyle\frac{d_2^2}{(4.37)^2}\Rightarrow\\\\\Rightarrow d_2^2=\displaystyle\frac{0.006329728*(4.37)^2}{50}\Rightarrow d_2^2=0.002417564\Rightarrow\\\\\Rightarrow d_2=\sqrt{0.002417564}=\boxed{0.049168726\;ly}$

Remark: It is worth noticing that Alpha Centauri A, though is the nearest star to the Sun, is not visible to the naked eye.

7 0

3 years ago

Howie works at a petting zoo. He fed a piglet 1/5 of a bottle of milk then gave 3/4 of what was left to the calf. How much of th

Lapatulllka [165]

Answer:

$\frac{3}{5}$ th part of the bottle will be drunk by the calf.

Step-by-step explanation:

Part of the bottle fed by the piglet = $\frac{1}{5}$

Milk left in the bottle = 1 - $\frac{1}{5}$

= $\frac{5-1}{5}$

= $\frac{4}{5}$

Milk drunk by the calf = $\frac{3}{4}$ th of the left part

= $\frac{3}{4}\times \frac{4}{5}$

= $\frac{3}{5}$ th part of the bottle

Therefore, $\frac{3}{5}$ th part of the bottle will be drunk by the calf.

7 0

3 years ago