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ASHA 777 [7]
3 years ago
7

A total of 32 students sign up for a kayaking trip. Four more seniors (s) sign up than juniors (j). Which system of linear equat

ions models this situation? A) s + j = 32 4s + j = 36 B) s + j = 32 s − j = 4 C) s + j = 36 s − j = 32 D) s + j = 36 −s − j = 4
Mathematics
1 answer:
horsena [70]3 years ago
7 0

Answer: OPTION B

Step-by-step explanation:

Let be "s" the number of seniors that sign up for the kayaking trip and "j" the number of juniors that sign up for the kayaking trip.

According to the information provided in the exercisse, a total of 32 students sign up; this means that the sum of the number of seniors and the number of juniors that sign up for the kayaking trip is 32. Then, you can write the following equation:

s+j=32

"Four more seniors sign up than juniors" indicates that the number of seniors that sign up is the number of juniors plus 4:

s=j+4

Subtracting "j" from both sides, you get:

s-j=j+4-j\\\\s-j=4

Therefore, the system of equations that model this situation is:

\left \{ {{s+j=32} \atop {s-j=4}} \right.

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I need to know the improper fractions answers for:
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Answer:

Part 1) x=\frac{15}{2}\ units

Part 2) z=\frac{15\sqrt{3}}{2}\ units

Part 3) y= \frac{15\sqrt{3}}{4}\ units

Part 4) b= \frac{45}{4}\ units

Step-by-step explanation:

step 1

Find the value of x

In the large right triangle

cos(60^o)=\frac{x}{15} ----> by CAH (adjacent side divided by the hypotenuse)

Remember that

cos(60^o)=\frac{1}{2}

substitute

\frac{1}{2}=\frac{x}{15}

solve for x

x=\frac{15}{2}\ units ---> improper fraction

step 2

Find the value of z

In the large right triangle

Applying the Pythagorean Theorem

15^2=x^2+z^2

substitute the value of x

15^2=(\frac{15}{2})^2+z^2

solve for z

z^2=15^2-(\frac{15}{2})^2

z^2=225-\frac{225}{4}

z^2=\frac{675}{4}

z=\frac{\sqrt{675}}{2}\ units

simplify

z=\frac{15\sqrt{3}}{2}\ units

step 3

Find the value of y

In the right triangle of the right

sin(30^o)=\frac{y}{z} ---> by SOH (opposite side divided by the hypotenuse)

substitute the given values of y and z

Remember that

sin(30^o)=\frac{1}{2}

so

\frac{1}{2}=y:\frac{15\sqrt{3}}{2}

solve for y

\frac{1}{2}= \frac{2y}{15\sqrt{3}}

y= \frac{15\sqrt{3}}{4}\ units

step 4

Find the value of b

In the right triangle of the right

cos(30^o)=\frac{b}{z} ---> by CAH (adjacent side divided by the hypotenuse)

substitute the given values of y and z

Remember that

cos(30^o)=\frac{\sqrt{3}}{2}

so

\frac{\sqrt{3}}{2}=b:\frac{15\sqrt{3}}{2}

solve for y

\frac{\sqrt{3}}{2}= \frac{2b}{15\sqrt{3}}

b= \frac{45}{4}\ units

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Given:

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