C= chair cost
t= table cost

Create two equations with the given information. Solve for one variable in equation one. Substitute that answer in equation two. Then you can solve for the needed information.
3c+2t=$18
5c+6t=$48

3c+2t=18
Subtract 2c from both sides
3c=18-2t
Divide both sides by 3
c=(18-2t)/3

Substitute the value for c in equation two:
5c+6t=$48
5((18-2t)/3)+6t=48
(90-10t)/3+6t=48
Multiply everything by 3 to eliminate fraction
(3)((90-10t)/3)+(3)(6t)=(3)(48)
90-10t+18t=144
90+8t=144
Subtract 90 from both sides
8t=54
Divide both sides by 8
t=$6.75 cost for table

Substitute the t value to solve for c:
3c+2t=18
3c+2(6.75)=18
3c+13.50=18
3c=4.50
c=$1.50 chair cost

Check:
5c+6t=$48
5(1.50)+6(6.75)=48
7.50+40.50=48
48=48

Hope this helps! :) If it does, please mark as brainliest.

7 0

3 years ago

Find the value of the six trig functions given the triangle below.

Y_Kistochka [10]

$\text{Here,}\\\\\text{Opposite side,}~ O = 2\\\\\text{Hypotenuse,}~ H = \sqrt 5 \\\\\text{Adjacent side,}~ A = \sqrt{5-4} = \sqrt 1 =1 ~~ ;[\text{By using Pythagorean theorem}]\\\\\\\sin \theta = \dfrac{O}H = \dfrac 2{\sqrt 5}\\\\\\\cos \theta = \dfrac{A}{H} = \dfrac{1}{\sqrt 5} \\ \\\\ \tan \theta = \dfrac{O}{A} = \dfrac 21 = 2$

$csc \theta = \dfrac{1}{\sin \theta} = \dfrac{\sqrt 5}2\\\\\\\sec \theta = \dfrac{1}{\cos \theta} = \sqrt 5 \\\\\\\cot \theta = \dfrac 1{\tan \theta} = \dfrac 12$

5 0

3 years ago

NEED HELP ASAP!! ...........

alexdok [17]

Answer:

i have no clue how to do thats,

5 0

2 years ago

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