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2 years ago

The price of a climbing trip to Mount Kilimanjaro includes an initial fee plus a constant fee per meter.

Mathematics

2 answers:

GenaCL600 [577]2 years ago

5 0

Answer:

<h2>1.44 per meter is the answer.. </h2>

Digiron [165]2 years ago

3 0

Answer:

u will get 1.44 per meter, if thats what ur looking for

Step-by-step explanation:

You might be interested in

A proportional relationship can be represented in the form of a

lidiya [134]

Answer:

a graph

Step-by-step explanation:

4 0

1 year ago

Nadia needs one cup of ice cream for every student in her class for their year-end party. The ice cream is only sold in pints. I

hoa [83]

The pints of ice cream for the students is an illustration of metric units

Nadia needs 12 pints of ice cream for the 24 students

<h3>How to determine the pints of ice cream?</h3>

The given parameter is:

Students = 24

She needs one cup for each student in the class.

So, the number of cups needed is:

Cups = 24

1 cup makes 1/2 pint.

So, we have:

Pint = 1/2 * 24

Pint = 12

Hence, Nadia needs 12 pints of ice cream for the 24 students

Read more about metric units at:

brainly.com/question/2004114

3 0

1 year ago

Can someone seriously help me!!!!!

igomit [66]

Answer:

Step-by-step explanation:

This should help I hope

3 0

3 years ago

In triangle ΔABC, ∠C is a right angle and CD is the height to

Zina [86]

Answer:

$m\angle CDB=90\\m\angle CBD=90-\alpha\\m\angle BCD=\alpha\\\\m\angle CDA=90\\m\angle CAD=\alpha\\m\angle ACD=90-\alpha$

Step-by-step explanation:

The triangles are drawn below.

CD is perpendicular to AB as CD is height to AB.

Therefore, angles $m\angle CDB=m\angle CDA=90$ °

So, triangles ΔCBD and ΔCAD are right angled triangles.

Now, from the right angled triangle ΔABC,

$m\angle A+m\angle B =90\\\alpha+m\angle B=90\\m\angle B=90-\alpha$

From ΔCBD,

$m\angle CBD$ is same as $m\angle B$ .

So, $m\angle CBD=90-\alpha$

$m\angle BCD+m\angle BDC =90\\m\angle BCD+90-\alpha=90\\m\angle BCD=\alpha$

Now, from ΔCAD,

$m\angle CAD$ is same as $m\angle A$

So, $m\angle CAD=\alpha$

$m\angle CAD+m\angle ACD =90\\\alpha+m\angle ACD=90\\m\angle ACD=90-\alpha$

Hence, the unknown angles of both the triangles are:

$m\angle CDB=90\\m\angle CBD=90-\alpha\\m\angle BCD=\alpha\\\\m\angle CDA=90\\m\angle CAD=\alpha\\m\angle ACD=90-\alpha$

5 0

2 years ago

How is breaking apart the problem 16x34 like solving four simpler problems?

dedylja [7]

10*30=300
10*4=40
6*30=180
6*4=24
add them up: 300+40+180+24=544

7 0

3 years ago