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valentina_108 [34]
3 years ago
14

im stupid so, if my grade was at a 90% (A) and my tests were worth 50% and i get a 40% on the test, what's my current grade now?

Mathematics
1 answer:
Butoxors [25]3 years ago
6 0
I think it would be a B
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Examine the first two steps used to solve the equation.
bija089 [108]

Answer:

C.

Step-by-step explanation:

Your fractions are missing there fraction bars:

3/4 (20y − 8) + 5 = 1/2 y + 1/4 (20y + 8)  

15y − 6 + 5 = 1/2 y + 5y + 2  

15y − 1 = 11/2 y + 2

A. 11/2y and 2 aren't like terms because one contains the variable y and the other contains no variable

B. The distribute property can't be used there because you don't have 15(y-1) you have 15y-1

C. Subtracting 11/2y sounds like a good step because there is a y term on the opposing side.

15y-1=11/2y+2

Subtracing 11/2y on both sides

9.5y-1=2

That looks pretty good because then you would add 1 on both sides giving:

9.5y =3

Last step would get the y by itself which is dividing both sides by 9.5 giving you 6/19.

D. You could actually do this but it doesn't help you get x by itself.  The equation would look like this: 15/2 y-1/2=11/4 y+1

5 0
3 years ago
Read 2 more answers
Write the phrase "9 increased by a number T" as an algebraic expression.
mars1129 [50]

Answer:

ok

Step-by-step explanation:

6 0
3 years ago
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I NEED HELP PLEASE !!!! :(
Triss [41]

Answer:

6

Step-by-step explanation:

(2)/((6)√(8))*√(2)-((18)/√81))-2)

(6)sqrt√8 = 6sqrt √2³= 2^3/6=2^1/2= √2

(2√2)/√2 -(-18/√9²)-2)

2-(-18/9 -2)

2-(-2-2)=2+4=6

4 0
3 years ago
Reparametrize the curve with respect to arc length measured from the point where t = 0 in the direction of increasing t. (Enter
jolli1 [7]

Answer:

the answer is incomplete, below is the complete question

"Reparametrize the curve with respect to arc length measured from the point where t = 0 in the direction of increasing t. (Enter your answer in terms of s.) r(t) = 3ti + (1 - 4t)j + (1 + 2t)k r(t(s)) ="

answer

r(t(s)) = \frac{3s}{\sqrt{29} } i + (1 -\frac{4s}{\sqrt{29} }t)j + (1 + \frac{2s}{\sqrt{29} })k

Step-by-step explanation:

The step by step procedure is to first determine the differentiate the given vector function

r(t) = 3ti + (1 - 4t)j + (1 + 2t)k

\frac{d(r(t) = 3ti + (1 - 4t)j + (1 + 2t)k)}{dt} \\r'(t)=3i-4j+2k\\

since s(t) is the arc length for r(t), which is define as

s(t)=\int\limits^t_0 {||r'(t)||} \, dt

if we substitute the value of r'(t) we arrive at

s(t)=\int\limits^t_0 {||r'(t)||} \, dt\\s(t)=\int\limits^t_0 {\sqrt{3^{2} +4^{2}+2^{2}} \, dt\\s(t)=\int\limits^t_ 0{\sqrt{29} } \, dx\\

s(t)=\int\limits^t_ 0{\sqrt{29} } \, dx\\\\s(t)=\sqrt{29} t\\hence \\t(s)=\frac{s}{\sqrt{29} }

substituting the value of t in to the given vector equation we have

r(t(s)) = \frac{3s}{\sqrt{29} } i + (1 -\frac{4s}{\sqrt{29} }t)j + (1 + \frac{2s}{\sqrt{29} })k

4 0
4 years ago
A manufacturer knows that their items have a normally distributed length, with a mean of 10.2 inches, and standard deviation of
ArbitrLikvidat [17]

Answer:

P [ X ≤  9.8 ]  = 0.1335

Step-by-step explanation:

P [ X ≤  9.8 ]  =  [ ( 9.8  - 10.2 )/1.8√25 ]

P [ X ≤  9.8 ]  =  - 0.4*5/1.8

P [ X ≤  9.8 ]  = - 2 / 1.8

P [ X ≤  9.8 ]  =  - 1.11

From z- table we get:  α  =  0.1335

P [ X ≤  9.8 ]  = 0.1335   or   P [ X ≤  9.8 ]  = 0.1335

8 0
3 years ago
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