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grandymaker [24]
2 years ago
8

Ms. Corley wants to take her class on a trip to either the nature center or the zoo the nature center charges $4 per student plu

s $95 for a one hour naturalist program. The zoo charges $9 per student plus $75 for a 1 hour tour guide
A. Solve the system of equations to represent this situation

B. Solve the system of equations algebraically. Interpret the solution.

C. Ms. Corley has 22 students in her class. Determine which would cost less, the nature center or the zoo


PLZZZ HELP!!
Mathematics
2 answers:
Tasya [4]2 years ago
7 0

Answer:

Step-by-step explanation:

Ms. Corley wants to take her class on a trip to either the nature center or the zoo.

Let the charge for the trip is y and number of student in the class be x.

For Nature center ⇒

Nature center charges $4 per students plus $95 for an hour.

Cost of the trip y = 4x + 95 ----------(1)

For Zoo ⇒

Zoo charges $9 per student plus $25 for one hour

Cost of the trip  y = 9x + 75 ------------(2)

(A) To solve the system of equations we will equate the value of y from equation (1) and (2)

y = 4x + 95 = 9x + 75

9x - 4x = 95 - 75

5x = 20

x = 4

and y = 4 × 4 + 95

         = 16 + 95

         = 111

(B) If Ms. Corley's class has 4 students then cost of the trip will be $111

(C) Cost of the trip to nature center = 4 × 22 + 95

                                                          = 88 + 95 = $183

Cost of the trip to the zoo = 9 × 22 + 75 = 198 + 75

                                             = $273

Therefore, cost of trip to the Nature center will be less.

lianna [129]2 years ago
3 0
X- class members
y-total cost of a trip
The nature center: y=4$*x+95$
The zoo: y=9$*x+75$
4x+95=9x+75⇒5x=20⇒x=20/5
x=4, y=111
<span>If Ms. Corley has 22 students in her class, cost for nature center would be 4$*22+95$=183$ and for the zoo 9$*22+75$=273$.
The zoo would cost less.
</span>
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Marty is asked to draw triangles with side lengths of 4 units and 2 units, and a non-included angle of 30°. Select all the trian
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Answer:

The drawn in the attached figure

see the explanation

Step-by-step explanation:

<em>First case</em>

In the triangle ABC

Let

a=4\ units\\b=2/ units\\B=30^o

Applying the law of sines

Find the measure of angle A

\frac{a}{sin(A)}=\frac{b}{sin(B)}

substitute the given values

\frac{4}{sin(A)}=\frac{2}{sin(30^o)}

sin(A)=1

so

A=90^o

Find the measure of angle C

In a right triangle

we know that

B+C=90^o ----> by complementary angles

B=30^o

therefore

C=60^o

Find the length side c

Applying the law of sines

\frac{c}{sin(C)}=\frac{b}{sin(B)}

substitute the given values

\frac{c}{sin(60^o)}=\frac{2}{sin(30^o)}

c=2\sqrt{3}\ units

therefore

The dimensions of the triangle are

A=90^o

B=30^o

C=60^o

a=4\ units\\b=2\ units\\c=2\sqrt{3}=3.46\ units

<em>Second case</em>

In the triangle ABC

Let

a=4\ units\\b=2/ units\\A=30^o

Applying the law of sines

Find the measure of angle B

\frac{a}{sin(A)}=\frac{b}{sin(B)}

substitute the given values

\frac{4}{sin(30^o)}=\frac{2}{sin(B)}

sin(B)=0.25

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B=14.48^o

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\frac{c}{sin(C)}=\frac{a}{sin(A)}

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A=30^o

B=14.48^o

C=135.52^o

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see the attached figure to better understand the problem

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2 years ago
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