1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
chubhunter [2.5K]
3 years ago
9

Nadia is mountain climbing. She started at an altitude of 19.26 feet below sea level and then changed her altitude by climbing a

total of 5,437.8 feet up from her initial position. What was Nadia's altitude at the end of her climb?
Mathematics
1 answer:
lesya692 [45]3 years ago
3 0

Answer:

5,418.2 feet

Step-by-step explanation:

Nadia is carrying out mountain climbing.

She started climbing the mountain at an altitude of 19.26 feet below the sea level.

Nadia changed her altitude by climbing a total of 5,437.8 feet from her starting position.

Therefore, Nadia's altitude at the end of her climb can be calculated as follows

= 5,437.8-19.6

= 5,418.2

Hence Maria's altitude at the end of her climb is 5,418.2 feet

You might be interested in
22. The measures of the angles of ADEF are in the ratio 2:4:4. What are the
denis23 [38]

The measures of the angles of Δ DEF are 36° , 72° , 72°

Step-by-step explanation:

The sum of the measures of the interior angles in any triangle is 180°

In Δ DEF

The ratio between the measures of its interior angles is 2 : 4 : 4

To find the measure of each angle

1. Find the sum of the ratio between the measures of its angles

2. Divide 180° by the sum of the ratio

3. Multiply the quotient by each ratio

∵ The ratio between the measures of the angles of Δ DEF = 2 : 4 : 4

∴ The sum of the ratio = 2 + 4 + 4 = 10

→    m∠D    :    m∠E    :    m∠F    :   sum of the ratio

→    2          :    4          :    4          :   10

→    ?          :     ?          :    ?          :   180°

Divide 180 by 10, then multiply the quotient by the components of

the ratio to find the measure of each angle

∵ m∠D = (2)\frac{180}{10}

∴ m∠D = 36°

∵ m∠E = (4)\frac{180}{10}

∴ m∠D = 72°

∵ m∠F = (4)\frac{180}{10}

∴ m∠F = 72°

The measures of the angles of Δ DEF are 36° , 72° , 72°

Learn more:

You can learn more about the ratio in brainly.com/question/2707032

#LearnwithBrainly

3 0
3 years ago
How can you tell if the solution is one value,all real numbers,or no solutions
zysi [14]
One value: would be a single valued function, or just one answer.

Real numbers would be: natural numbers, whole numbers, integers, rational numbers (fractions and repeating or terminating decimals), and irrational numbers.

No solution: there is no answer to the question.
8 0
3 years ago
Somone please help me asap i need to turn this in its due tomorrow ​
Novosadov [1.4K]
Hope this helps !
Answer circled on green

6 0
2 years ago
Find the area of the shape shown below.<br> 12<br> 5<br> 5<br> units
Yanka [14]

Answer:

42.5 units²

Explenation:

The area of the square on the left is 5×5=25.

At the top you do 12-5=7 to find the length of the tringle.

Than it is 7×5=35 than this divided by two, so 17.5.

You add both areas 25+17.5=42.5

5 0
2 years ago
Read 2 more answers
Which of the following is not the graph of a function of x?​
Paraphin [41]

Answer:

i think it's the third one

6 0
3 years ago
Other questions:
  • Round 14102.60942 to 3sf
    14·1 answer
  • Factorise 7m^2+6m-1<br> Thanks
    15·1 answer
  • Can someone help me with this?
    7·2 answers
  • What number should be added to the expression x^2 + 6x to change it into a perfect square trinomal
    15·1 answer
  • HAlp meh thank you &gt;:3
    6·2 answers
  • Select the perfect square trinomial from among the options below!
    7·2 answers
  • Pls help find x in both
    12·1 answer
  • Literal Equations:
    7·1 answer
  • A 24 foot tall streetlight casts a shadow that is 18 feet long. How long of a shadow is cast by a nearby parking meter post that
    12·2 answers
  • I NEED HELP FAST!!!<br> What is the distance between 14,-15 and 10,-7
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!