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Natasha2012 [34]
3 years ago
11

Rebecca makes an A in a 4-credit math course , a B in a 3-credit English course, a C in a 3-credit science course, a B in the as

sociated 1-credit science lab, and an A in a 4-credit Spanish course. What is her GPA for the semester? A- 1.067 B- 3.2 C- 3.333 D- 6.667
Mathematics
1 answer:
Elan Coil [88]3 years ago
4 0

Answer: b

Step-by-step explanation: krusty crab pizza is the pizza for you and me

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What is the scale factor from Figure 2 to Figure 1?
Luda [366]

Answer:

3

Step-by-step explanation:

Each of the lines lengths multiply by 3.

6 0
2 years ago
A^2(64a^2-3)/3=27-4a^2/4
frutty [35]
The answer is a = 3/4 = 0.75

First get rid of the paranthesis,
{a}^{2} (64 {a}^{2}  - 3) = 64 {a}^{4} -  3 {a}^{2}
Then set the denominators equal:
\frac{ 256 {a}^{4}  - 12 {a}^{2}}{12}  =  \frac{81 - 12 {a}^{2} }{12}
Then remove the denominators and solve:
256 {a}^{4}  - 12 {a}^{2}  = 81 - 12 {a}^{2}
Eliminate -12a^2 by adding 12a^2 to both sides:
256 {a}^{4}  = 81
Take the fourth root of them or take the square root twice:
\sqrt[4]{256 {a}^{4} }  =  \sqrt[4]{81}   \\  4a = 3
Divide both sides by 4:
a =  \frac{3}{4}  = 0.75
7 0
2 years ago
ASAP PLEASE HELP But he circled just beyond the range of the club, snarling with bitterness and rage; and while he circled he wa
vredina [299]

Answer: Your answer is A!

Step-by-step explanation:

5 0
2 years ago
Solve for x x squared plus 4X minus 21 equals 0
suter [353]

Answer:

x = -7 and x = 3

Step-by-step explanation:

x² + 4x - 21 = 0 factors as follows:  (x + 7)(x - 3) = 0.

Then x = -7 and x = 3.

3 0
3 years ago
Read 2 more answers
Evaluate the integral ∫2−1|x−1|dx
defon

I think you might be referring to the definite integral,

\displaystyle \int_{-1}^2|x-1|\,\mathrm dx

Recall the definition of absolute value:

|x| = \begin{cases}x&\text{if }x\ge0\\-x&\text{if }x

Then |x-1|=x-1 if x\ge1, and |x-1|=1-x is x. So spliting up the integral at <em>x</em> = 1, we have

\displaystyle \int_{-1}^2|x-1|\,\mathrm dx = \int_{-1}^1(1-x)\,\mathrm dx + \int_1^2(x-1)\,\mathrm dx

The rest is simple:

\displaystyle \int_{-1}^2|x-1|\,\mathrm dx = \left(x-\dfrac{x^2}2\right)\bigg|_{-1}^1 + \left(\dfrac{x^2}2-x\right)\bigg|_1^2 \\\\ = \left(\left(1-\frac12\right)-\left(-1-\frac12\right)\right) + \left(\left(2-2\right)-\left(\frac12-1\right)\right) \\\\ = \boxed{\frac52}

5 0
2 years ago
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