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kow [346]
2 years ago
13

If a cow has a mass of 9×102 kilograms, and a blue whale has a mass of 1.8×105 kilograms, which of these statements is true?

Mathematics
1 answer:
lana66690 [7]2 years ago
7 0

Answer:

The mass of the Blue whale is 200 times the mass of the cow

Step-by-step explanation:

Given

Mass of Cow = 9 * 10² kg

Mass of Blue Whale = 1.8 * 10⁵ kg

Required

Determine the relationship between both weights

Represent the mass of the cow with C and the mass of the whale with B

C = 9 * 10^2kg

C = 9 *100kg

C = 900kg

B = 1.8 * 10^5kg

B = 1.8 * 100000kg

B = 180000kg

Divide the bigger weight by the smaller weight

\frac{B}{C} = \frac{180000kg}{900kg}

\frac{B}{C} = \frac{180000}{900}

\frac{B}{C} = {200}{}

Multiply both sides by C

C * \frac{B}{C} = {200}{} * C

B = {200}{} * C

B = 200}C

<em>From the expression above, it can be concluded that the mass of the Blue whale is 200 times the mass of the cow</em>

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Dakari makes 2 dozen jars of jam. The total weight of all of the filled jars is 211.2 ounces. When empty, each jar weighs 1.2 ou
Rzqust [24]

Answer:

Step-by-step explanation:

Total weight of all of the filled jars with jam = 211.2 ounces

When empty, each jar weighs 1.2 ounces

Total weight of 2 dozens of empty jar = 24 × 1.2 ounces

= 28.8 ounces

Total weight of jam in 2 dozens jar = Total weight of all of the filled jars with jam - Total weight of 2 dozens of empty jar

= 211.2 ounces - 28.8 ounces

= 182.4 ounces

The weight of jam in one jar = 182.4 ounces / 24 jars

= 7.6 ounces

The weight of jam in one jar = 7.6 ounces

3 0
2 years ago
What is the simplest form of RootIndex 3 StartRoot 27 a cubed b Superscript 7 Baseline EndRoot?
Vinvika [58]

Answer: 3ab\sqrt[3]{b^4}

Step-by-step explanation:

Given the following expression:

\sqrt[3]{27a^3b^7}

You need to apply the Product of powers property, which states that:

(a^m)(a^n)=a^{(m+n)

Then, you can rewrite the expression as following:

=\sqrt[3]{27a^3b^4b^3}

The next step is to descompose 27 into its prime factors:

27=3*3*3=3^3

Now you must substitute 3^3 inside the given root. Then:

=\sqrt[3]{3^3a^3b^4b^3}

You need to remember that, according to Radicals properties:

\sqrt[n]{a^n}=a^{\frac{n}{n}}=a^1=a

Therefore, the final step is to apply this property in order to finally get the expression is its simplest form. This is:

=3^{\frac{3}{3}}a^{\frac{3}{3}}b^{\frac{4}{3}}b^{\frac{3}{3}}=3ab^{\frac{4}{3}}b=3ab\sqrt[3]{b^4}

3 0
2 years ago
Read 2 more answers
Law of sines: StartFraction sine (uppercase A) Over a EndFraction = StartFraction sine (uppercase B) Over b EndFraction = StartF
vladimir2022 [97]

Answer:

30°

Step-by-step explanation:

Law of Sines = \frac{a}{Sin A} = \frac{b}{Sin B} = \frac{c}{Sin C}

\frac{2}{Sin 30} = \frac{2}{Sin B}

\frac{2}{Sin 30} = \frac{2}{Sin 30}

Therefore, m∠B = 30°

Hope that's right and helps

6 0
2 years ago
A County Superintendent of Highways is interested in the numbers of different types of vehicles that regularly travel within his
Karolina [17]

Answer:

a) The ratio of passenger cars to pickup trucks is \frac{5}{7}

The ratio of pickup trucks to passenger cars is \frac{7}{5}

The ratio of passenger cars to total vehicles is \frac{5}{12}

The ratio of pickup trucks to total vehicles is \frac{7}{12}

b) The tape diagram is shown in the attached picture.

c) The tape diagram has 12 equal-sized parts.

d) The total quantity that the tape diagram represent is 192.

e) The value of each individual part of the tape is 16.

f) There were registered 80 passenger cars and 112 pickup trucks.

Step-by-step explanation:

a) According to the information in the problem, that for every 5 passenger cars registered, there were 7 pickup trucks registered.

So the ratios are:

Passenger cars to pickup trucks: \frac{5}{7}

Pickup trucks to passenger cars: \frac{7}{5}

If we calculate the total vehicles: 5+7=12

So, we can get two ratios more:

Passenger cars to total vehicles is \frac{5}{12}

Pickup trucks to total vehicles is \frac{7}{12}

b) We make a tape diagram where each box represents a vehicle so we draw 5 boxes for cars and 7 boxes for pickup trucks.

c) If we count the total quantity of boxes, there are 12.

d) This diagram represents the total quantity of registrations in August which is 192.

e) To get the representation of each part of the tape diagram we have to divide 192 by 12. 192÷12=16

f) Finally, we multiply 16 by the number of boxes of each type of vehicle:

Cars: 16×5=80

Pickup trucks: 16×7=112

6 0
2 years ago
Three erasers and five pencils cost $7.55. 6 erasers and 12 pencils cost $17.40. How much does 1 eraser cost? How much does 1 pe
Sphinxa [80]

Answer:

Step-by-step explanation:

erasers=e  

pencils=p

3e+5p=7.55   ...(1)

6e+12p=17.40

divide by 2

3e+6p=8.70  ...(2)

(2)-(1) gives

p=8.70-7.55=1.15

from (1)

3e+5(1.15)=7.55

3e+5.75=7.55

3e=7.55-5.75

3e=1.80

e=1.80/3=0.60

cost of 1 eraser=$0.60

cost of 1 pencil =$1.15

7 0
2 years ago
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