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Citrus2011 [14]
3 years ago
13

Write an equation or inequality to solve this problem. On three extra credit problems on a math test, students are given up to 4

points each, provided that their scores do not exceed 100 (points). What is the minimum raw score students could have and still score 100 with the extra credit?
Mathematics
2 answers:
Zepler [3.9K]3 years ago
8 0
3 problems worth up to 4 points each means 12 possible points
x+12=100
x=88
vlada-n [284]3 years ago
4 0
<span>100-12 points=88
88 is the maximum raw score possible..

: 3</span>
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PLEASEEEEE HELPPPP ONLY NUMBER 2 !!!!!<br> Due todayyy!!!<br> 10 pointssss
forsale [732]

Answer: Answer B maybe?

Step-by-step explanation: I'm in sixth grade, so i don't know, but I would guess its answer B, if i'm correct, i have a 4.000 G.P.A so i hope that helps?

5 0
3 years ago
Read 2 more answers
Find the projection of the vector A = î - 2ġ + k on the vector B = 4 i - 4ſ + 7k. 15. Given the vectors A = 2 i +3 ſ +6k and B =
Gwar [14]

Answer:

Part 1)

Projection of vector A on vector B equals 19 units

Part 2)

Projection of vector B' on vector A' equals 35 units

Step-by-step explanation:

For 2 vectors A and B the projection of A on B is given by the vector dot product of vector A and B

Given

\overrightarrow{v_{a}}=\widehat{i}-2\widehat{j}+\widehat{k}

Similarly vector B is written as

\overrightarrow{v_{b}}=4\widehat{i}-4\widehat{j}+7\widehat{k}

Thus the vector dot product of the 2 vectors is obtained as

\overrightarrow{v_{a}}\cdot \overrightarrow{v_{b}}=(\widehat{i}-2\widehat{j}+\widehat{k})\cdot (4\widehat{i}-4\widehat{j}+7\widehat{k})\\\\\overrightarrow{v_{a}}\cdot \overrightarrow{v_{b}}=1\cdot 4+2\cdot 4+1\cdot 7=19

Part 2)

Given vector A' as

\overrightarrow{v_{a'}}=2\widehat{i}+3\widehat{j}+6\widehat{k}

Similarly vector B' is written as

\overrightarrow{v_{b'}}=\widehat{i}+5\widehat{j}+3\widehat{k}

Thus the vector dot product of the 2 vectors is obtained as

\overrightarrow{v_{b'}}\cdot \overrightarrow{v_{a'}}=(\widehat{i}+5\widehat{j}+3\widehat{k})\cdot (2\widehat{i}+3\widehat{j}+6\widehat{k})\\\\\overrightarrow{v_{a'}}\cdot \overrightarrow{v_{b'}}=1\cdot 2+5\cdot 3+3\cdot 6=35

7 0
4 years ago
Combine like terms<br><br> 4y^2+4(7y^2-8)=<br><br> please show your work
kirza4 [7]

Answer:

32y^2-32

Step-by-step explanation:

4y^2+4(7y^2-8)

4y^2+28y^2-32

32y^2-32

if you need extra steeps explaining then plz let me know and I will help :)

8 0
3 years ago
Please help I will give Brainliest please!
WITCHER [35]

Part (a)

The domain is the set of allowed x inputs of a function.

The graph shows that x = 0 is not allowed because of the vertical asymptote located here. It seems like any other x value is fine though.

<h3>Domain: set of all real numbers but x \ne 0</h3>

To write this in interval notation, we can say (-\infty, 0) \cup (0, \infty) which is the result of poking a hole at 0 on the real number line.

--------------

The range deals with the y values. The graph makes it seem like it stretches on forever in both up and down directions. If this is the case, then the range is the set of all real numbers.

<h3>Range: Set of all real numbers</h3>

In interval notation, we would say (-\infty, \infty) which is almost identical to the interval notation of the domain, except this time of course we aren't poking at hole at 0.

=======================================================

Part (b)

<h3>The x intercepts are x = -4 and x = 4</h3>

We can compact that to the notation x = \pm 4

These are the locations where the blue hyperbolic curve crosses the x axis.

=======================================================

Part (c)

<h3>Answer: There aren't any horizontal asymptotes in this graph.</h3>

Reason: The presence of an oblique asymptote cancels out any potential for a horizontal asymptote.

=======================================================

Part (d)

The vertical asymptote is located at x = 0, so the equation of the vertical asymptote is naturally x = 0. Every point on the vertical dashed line has an x coordinate of zero. The y coordinate can be anything you want.

<h3>Answer: x = 0 is the vertical asymptote</h3>

=======================================================

Part (e)

The oblique or slant asymptote is the diagonal dashed line.

It goes through (0,0) and (2,6)

The equation of the line through those points is y = 3x

If you were to zoom out on the graph (if possible), then you should notice the branches of the hyperbola stretch forever upward but they slowly should approach the "fencing" that is y = 3x. The same goes for the vertical asymptote as well of course.

<h3>Answer:  Oblique asymptote is y = 3x</h3>
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3 years ago
Which number sequence follows the rule add 4 starting from 2?
Kaylis [27]
2 6 10 14 18 22 26 30 and so on
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2 years ago
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