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olasank [31]
3 years ago
13

If I have a GPA Of (2.6552) and makes straight A's My 3rd & 4th Quarter, what will be My graduating GPA?

Mathematics
2 answers:
IgorLugansk [536]3 years ago
5 0
Tell me all ur classes. Like Math, History etc.....Give me all 8 clases name

sveticcg [70]3 years ago
3 0
Can I have all subjects you do at school
You might be interested in
Kara bought two birthday gifts for her twin nephews. Each gift cost $12.50. Sales tax was
mixas84 [53]

Answer:

$27.12

Step-by-step:

So, we first need to learn how much money the tax is. We can find it like this

8.5 8.5x12.50

------ x 12.50 = ----------------

100 100

106.25

= ------------- = 1.0625 ~~ $1.06

100

The tax is $1.06

So now, the only thing left is to add

12.50 + 1.06 = $13.56

13.56x2 = $27.12 (for both gifts)

In total, both gifts cost $27.12 with tax.

In total, both gifts cost

8 0
3 years ago
Solve the equation for x.<br><br> 1<br> 4<br> (2x + 8) = −16
dimaraw [331]

Answer:Let's solve your equation step-by-step.

1

4

(2x+8)=−16

Step 1: Simplify both sides of the equation.

1

4

(2x+8)=−16

(

1

4

)(2x)+(

1

4

)(8)=−16(Distribute)

1

2

x+2=−16

Step 2: Subtract 2 from both sides.

1

2

x+2−2=−16−2

1

2

x=−18

Step 3: Multiply both sides by 2.

2*(

1

2

x)=(2)*(−18)

x=−36

4 0
3 years ago
Read 2 more answers
Identify a pair of supplementary angles
Elena-2011 [213]

Answer:

1st option

Step-by-step explanation:

6 0
4 years ago
Question 22 of 25<br> Which of the following is most likely the next step in the series?
forsale [732]

Answer:

B

Step-by-step explanation:

6 0
3 years ago
What is a cubic polynomial function in standard form with zeroes 1, –2, and 2?
Lisa [10]
Starting off with the polynomial in standard form would be extremely difficult, but we can construct one fairly easily with the zeroes we've been given.

We know from the given zeroes that our function has the value 0 when x = 1, x = -2, and x = 2. Manipulating each equation, we can rewrite them as x - 1 = 0, x + 2 = 0, and x - 2 = 0. To construct our polynomial, we simply use all three of the expressions on the left side of the equation as factors and multiply them together, obtaining:

(x-1)(x+2)(x-2)=0

Notice that we can easily obtain each our three zeroes by dividing both sides by the two other factors. From here, we just need to expand the left-hand side of the equation. I'll show the work required here:

(x-1)(x+2)(x-2)=0\\&#10;\big[(x-1)x+(x-1)2\big](x-2)=0\\&#10;(x^2-x+2x-2)(x-2)=0\\&#10;(x^2+x-2)(x-2)=0\\&#10;(x^2+x-2)x-(x^2+x-2)2=0\\&#10;x^3+x^2-2x-(2x^2+2x-4)=0\\&#10;x^3+x^2-2x-2x^2-2x+4=0\\&#10;x^3+(x^2-2x^2)+(-2x-2x)+4=0\\&#10;x^3-x^2-4x+4=0\\

So, in standard form, our cubic polynomial would be x^3-x^2-4x+4
3 0
4 years ago
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