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Evgesh-ka [11]
2 years ago
6

Jared and Casey splurged on a special dinner for their anniversary. Their dinner cost $262.57, and the local sales tax is 4%. If

they left a 20% tip on the $262.57, how much in total did they pay?
Mathematics
1 answer:
Maru [420]2 years ago
3 0

Answer:

325.58

Step-by-step explanation:

so im not good at explaining but JUST their DINNER costed $262.57. so you also have to ADD TAX because everything has tax so to find the 4% sale tax you need to set up the equation 262.57 x .04 (which is 4%). you will get 10.50 if you round. now you need to ADD TIP which is 20%. so to find that you need to do 262.57 x .20 (which is 20%). you will get 52.51. so now take BOTH 52.51 and 10.50 and ADD them. when you get this you also add it to the 262.57 and it gets you to 325.58

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You invest a total of $5800 in two investments earning 3.5% and 5.5% simple interest. Your goal is to have a total annual intere
attashe74 [19]
The annual interest that can be earned through investment of an amount at a simple interest can be calculated through the equation,
                I  = P x (i)
where I is interest, P is the principal amount, and i is the decimal equivalent of the interest. 

Let x be the amount deposited with 3.5% interest. With this representation, the amount deposited with 5.5% is 5800 - x. 

The linear equation that would represent the given scenario is,
        x(0.035) + (5800 - x)(0.055) = 283

Simplifying the equation,
 
       0.035x + 319 - 0.055x = 283

Combining like terms,
           -0.02x = -36
Dividing by -0.02,   
               x = 1800
             $5800 - x = $5800 - $1800 = y
                   y = $4000

Hence, the value that should be deposited to the 5.5% is $40000.   
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Which of the ordered pairs in the form (x,y)is a solution of this equation?
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Question 20 not yet answered marked out of 1.00 not flaggedflag question question text luna created a trash can in the shape of
Alex Ar [27]

Consider the triangle with sides a=1.6, b=2.3 and c=1.2 feet. You can use cosine theorem twice to find two unknown angles:

1.

b^2=a^2+c^2-2ac\cos \angle B,\\ (2.3)^2=(1.6)^2+(1.2)^2-2\cdot 1.6\cdot 1.2\cos \angle B,\\ 5.29=2.56+1.44-3.84\cos \angle B,\\ 3.84\cos \angle B=4-5.29=-1.29,\\\cos \angle B=-\dfrac{1.29}{3.84} =-0.336

Then you can conclude that ∠B is obtuse (because \cos \angle B) and m∠B=110°.

2.

c^2=a^2+b^2-2ab\cos \angle C,\\ (1.2)^2=(1.6)^2+(2.3)^2-2\cdot 1.6\cdot 2.3\cos \angle C,\\ 1.44=2.56+5.29-7.36\cos \angle C,\\ 7.36\cos \angle C=7.85-1.44=6.41,\\\cos \angle C=\dfrac{6.41}{7.36} =0.871.

Then you can conclude that ∠C is acute (because \cos \angle C>0) and m∠C=29°.

3. m∠A+m∠B+m∠C=180°, then m∠A=180°-110°-29°=41°.

Answer: m∠A=41° (∠A is opposite to the side a=1.6), m∠B=110° (∠B is opposite to the side b=2.3) and m∠C=29° (∠C is opposite to the side c=1.2)

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3 years ago
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