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leonid [27]
3 years ago
5

If i have a 83.5 as my grade and let's just say i would get a 70/100 what would be my grade?

Mathematics
1 answer:
daser333 [38]3 years ago
6 0

Answer:

76.75

Step-by-step explanation:

83.5 + 70=  153.5

153.5 / 2 = 76.75

Your grade would be 76.75.

HOPE THIS HELPED

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Car is driving at 30 kilometers per hour. How far in meters does it travel in 3 seconds?
Alenkasestr [34]

Answer:

25 meters

Step-by-step explanation:

30 km/hr = (30 km/1 hr)*(1000 m/1 km)*(1 hr/60 min)*(1 min/60 sec)

30 km/hr = (30*1000*1*1)/(1*1*60*60) meters per second

30 km/hr = 8.3333333333 m/s (approximate)

----------

The speed or rate is r = 8.3333333333 meters per second approximately, and the time is t = 3 seconds, so,

distance = rate*time

d = r*t

d = 8.3333333333 * 3

d = 24.9999999999

Due to rounding error, the true answer should be d = 25. We can see this if we opted to use the fraction from of 8.3333... instead of the decimal representation which isn't a perfect exact measurement.

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Find x. Assume that segments that appear tangent are tangent.
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Please help solve these proofs asap!!!
timama [110]

Answer:

Proofs contained within the explanation.

Step-by-step explanation:

These induction proofs will consist of a base case, assumption of the equation holding for a certain unknown natural number, and then proving it is true for the next natural number.

a)

Proof

Base case:

We want to shown the given equation is true for n=1:

The first term on left is 2 so when n=1 the sum of the left is 2.

Now what do we get for the right when n=1:

\frac{1}{2}(1)(3(1)+1)

\frac{1}{2}(3+1)

\frac{1}{2}(4)

2

So the equation holds for n=1 since this leads to the true equation 2=2:

We are going to assume the following equation holds for some integer k greater than or equal to 1:

2+5+8+\cdots+(3k-1)=\frac{1}{2}k(3k+1)

Given this assumption we want to show the following:

2+5+8+\cdots+(3(k+1)-1)=\frac{1}{2}(k+1)(3(k+1)+1)

Let's start with the left hand side:

2+5+8+\cdots+(3(k+1)-1)

2+5+8+\cdots+(3k-1)+(3(k+1)-1)

The first k terms we know have a sum of .5k(3k+1) by our assumption.

\frac{1}{2}k(3k+1)+(3(k+1)-1)

Distribute for the second term:

\frac{1}{2}k(3k+1)+(3k+3-1)

Combine terms in second term:

\frac{1}{2}k(3k+1)+(3k+2)

Factor out a half from both terms:

\frac{1}{2}[k(3k+1)+2(3k+2]

Distribute for both first and second term in the [ ].

\frac{1}{2}[3k^2+k+6k+4]

Combine like terms in the [ ].

\frac{1}{2}[3k^2+7k+4

The thing inside the [ ] is called a quadratic expression.  It has a coefficient of 3 so we need to find two numbers that multiply to be ac (3*4) and add up to be b (7).

Those numbers would be 3 and 4 since

3(4)=12 and 3+4=7.

So we are going to factor by grouping now after substituting 7k for 3k+4k:

\frac{1}{2}[3k^2+3k+4k+4]

\frac{1}{2}[3k(k+1)+4(k+1)]

\frac{1}{2}[(k+1)(3k+4)]

\frac{1}{2}(k+1)(3k+4)

\frac{1}{2}(k+1)(3(k+1)+1).

Therefore for all integers n equal or greater than 1 the following equation holds:

2+5+8+\cdots+(3n-1)=\frac{1}{2}n(3n+1)

//

b)

Proof:

Base case: When n=1, the left hand side is 1.

The right hand at n=1 gives us:

\frac{1}{4}(5^1-1)

\frac{1}{4}(5-1)

\frac{1}{4}(4)

1

So both sides are 1 for n=1, therefore the equation holds for the base case, n=1.

We want to assume the following equation holds for some natural k:

1+5+5^2+\cdots+5^{k-1}=\frac{1}{4}(5^k-1).

We are going to use this assumption to show the following:

1+5+5^2+\cdots+5^{(k+1)-1}=\frac{1}{4}(5^{k+1}-1)

Let's start with the left side:

1+5+5^2+\cdots+5^{(k+1)-1}

1+5+5^2+\cdots+5^{k-1}+5^{(k+1)-1}

We know the sum of the first k terms is 1/4(5^k-1) given by our assumption:

\frac{1}{4}(5^k-1)+5^{(k+1)-1}

\frac{1}{4}(5^k-1)+5^k

Factor out the 1/4 from both of the two terms:

\frac{1}{4}[(5^k-1)+4(5^k)]

\frac{1}{4}[5^k-1+4\cdot5^k]

Combine the like terms inside the [ ]:

\frac{1}{4}(5 \cdot 5^k-1)

Apply law of exponents:

\frac{1}{4}(5^{k+1}-1)

Therefore the following equation holds for all natural n:

1+5+5^2+\cdots+5^{n-1}=\frac{1}{4}(5^n-1).

//

5 0
3 years ago
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