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kari74 [83]
3 years ago
15

I have 4 times the homework than my brother. I have 9 more sheets than him. How many sheets of homework do we have altogether?

Mathematics
1 answer:
dem82 [27]3 years ago
7 0

Answer:

15 sheets of homework

Step-by-step explanation:

That is because you have 12 sheets of homework and your brother has 3 sheets of homework. 3*4=12 and 3+9=12

So 12 + 3 =15

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Find m∠U.<br> Write your answer as an integer or as a decimal rounded to the nearest tenth.
Elodia [21]

Answer:

56.4°

Step-by-step explanation:

sin(5/6)

= 0.98511 rad = 56.443°

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3 years ago
What is the value of x in the solution to the system of linear equations?
xenn [34]
Since y=x-4 you would simply just plug that into 'y' on the other equation, given: x-4=3x+2.
Subtract x from both sides, it will leave you with -4=2x+2.
Next you would subtract 2 from both sides, leaving you with -6=2x.
All you need now is to get rid of the coefficient in front of 'x' so you divide both sides with 2, leaving you with your answer: -3=x.

Answer: -3=x

Hope this helps!
6 0
3 years ago
Read 2 more answers
Gilbert is graduating from college in one year, but he will need a loan in the amount of $5,125 for his last two semesters. He m
sleet_krkn [62]
First, determine the effective interests given both interest rates.

(1)   ieff = (1 + 0.068/12)^12 - 1 = 0.07016
(2)   ieff = (1 + 0.078/12)^12 - 1 = 0.08085

Calculating the interests will entail us to use the equation,
     I = P ((1 + i)^n - 1)

Substituting the known values,
  (1)    I = ($5125)((1 + 0.07016)^1/2 - 1)
          I = $176.737

   (2)    I = ($5125)(1 + 0.08085)^1/2 - 1)
            I = $203.15

a. Hence, the greater interest will be that of the second loan. 

b. The difference between the interests,
    d = $203.15 - $176.737
          $26.413
7 0
3 years ago
Use triangle ABC drawn below &amp; only the sides labeled. Find the side of length AB in terms of side a, side b &amp; angle C o
Brrunno [24]

Answer:

AB = \sqrt{a^2 + b^2-2abCos\ C}

Step-by-step explanation:

Given:

The above triangle

Required

Solve for AB in terms of a, b and angle C

Considering right angled triangle BOC where O is the point between b-x and x

From BOC, we have that:

Sin\ C = \frac{h}{a}

Make h the subject:

h = aSin\ C

Also, in BOC (Using Pythagoras)

a^2 = h^2 + x^2

Make x^2 the subject

x^2 = a^2 - h^2

Substitute aSin\ C for h

x^2 = a^2 - h^2 becomes

x^2 = a^2 - (aSin\ C)^2

x^2 = a^2 - a^2Sin^2\ C

Factorize

x^2 = a^2 (1 - Sin^2\ C)

In trigonometry:

Cos^2C = 1-Sin^2C

So, we have that:

x^2 = a^2 Cos^2\ C

Take square roots of both sides

x= aCos\ C

In triangle BOA, applying Pythagoras theorem, we have that:

AB^2 = h^2 + (b-x)^2

Open bracket

AB^2 = h^2 + b^2-2bx+x^2

Substitute x= aCos\ C and h = aSin\ C in AB^2 = h^2 + b^2-2bx+x^2

AB^2 = h^2 + b^2-2bx+x^2

AB^2 = (aSin\ C)^2 + b^2-2b(aCos\ C)+(aCos\ C)^2

Open Bracket

AB^2 = a^2Sin^2\ C + b^2-2abCos\ C+a^2Cos^2\ C

Reorder

AB^2 = a^2Sin^2\ C +a^2Cos^2\ C + b^2-2abCos\ C

Factorize:

AB^2 = a^2(Sin^2\ C +Cos^2\ C) + b^2-2abCos\ C

In trigonometry:

Sin^2C + Cos^2 = 1

So, we have that:

AB^2 = a^2 * 1 + b^2-2abCos\ C

AB^2 = a^2 + b^2-2abCos\ C

Take square roots of both sides

AB = \sqrt{a^2 + b^2-2abCos\ C}

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3 years ago
Simplify <br> 2(10) + 2(x – 4)
Amanda [17]

Answer: 2x+12

Step-by-step explanation:

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8 0
3 years ago
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