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rusak2 [61]
3 years ago
10

Which statement is true about the function graphed here

Mathematics
1 answer:
Juli2301 [7.4K]3 years ago
5 0

Answer:

The y-intercept is (0,-6), the vertex is (4,2)

Step-by-step explanation:

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How do you write 680,705 in expanded form??
atroni [7]

<em>Hello!</em>

<em>The expanded form is:</em>

<em>600,000 + 80,000 + 700 + 5</em>


<em>Toodles~</em>

4 0
3 years ago
Which step will get you closest to the solution of this equation 5×(x-2)=7
scZoUnD [109]

Answer:

Division property get you to the quick solution

Solution: 3.4

Step-by-step explanation:

5*(x - 2) = 7

Dividing both sides by 5

5*(x -2)÷5 = 7÷5

x - 2 = 1.4

Adding 2 on both sides, we

x -2 + 2 = 1.4 + 2

x = 3.4

Hope you will understand the concept.

Thank you.

8 0
3 years ago
16 x + 4 y = − 6<br><br> 8 x + 2 y = 0
Vera_Pavlovna [14]
The answer is must be no solution
5 0
3 years ago
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
This is a late assignment and I would appreciate if someone could help me answer it quiiiiick! Thank you!
zhenek [66]
The answer is on the paper

5 0
3 years ago
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