Which could be the graph of y = -3(x + 1)2 + 2?

If F(x)=2x^2-30 find f(4)

valentinak56 [21]

Substitute 4 for x in the expression. After substituting for x, your new problem will be: 2(4)^2 - 30.

Solve the exponents first, so now you will have 2(16) - 30.

Multiply 2 times 16. 2 * 16 = 32, so now your problem should look like: 32 - 30.

Subtract to get your final answer of 2.

f(4) = 2

7 0

4 years ago

Graph the systems of equations.<br><br> {2x+y=8−x+2y=6

victus00 [196]

Hi, I actually just took the test and got 100%

Remember: When plotting the points for this equation, make sure to always first plot the ones that correspond to the first linear equation, and then plot the ones that correspond to the second linear equation.

The points on the line should be for the first linear equation, (4,0) and (8,0). I got this answer by first converting the linear equation, 2x+y=8 from standard form to slope-intercept form. To do this, I subtracted 2x from both sides of the equation. So now it reads as y=8-2x. After this step was completed, I then graphed my first linear equation.
The points on the line should be for the first linear equation, (2,4) and (6,6).
I got this answer by first converting the linear equation, -x+2y=6 into slope-intercept form. To do this, I subtracted -x from both sides of the equation. Then I had to divide the 2 into both -x and 6. So now it reads as y= 6/2-x/2. After this step was completed, I then graphed my second and final linear equation.

I hope this helps!

5 0

3 years ago

Read 2 more answers

Drag and drop a statement or reason to each box to complete the proof. Given: parallelogram EFGH Prove: EG¯¯¯¯¯ bisects HF¯¯¯¯¯¯

Archy [21]

Answer:

The question is incomplete. The complete question has been added as an attachment to the solution.

To aid easy identification of the unknowns, I attached another document where I numbered the unknowns.

1) EF ≅ HG

2) EF || HG

3) <FEK ≅ <HGK

<EFK ≅ <GHK

4) When two parallel lines are cut by a transversal, alternate interior angles are congruent.

5) EK ≅ GK

FK ≅ HK

Step-by-step explanation:

1) In the given parallelogram EFGH,

It is stated that the opposite sides of a parallelogram are congruent. Therefore, EF ≅ HG

2) It would be noted from the properties of a parallelogram, that the opposite sides are parallel. Therefore, EF || HG

3) Definition of a parallelogram: A parallelogram is a quadrilateral with two opposite sides that are parallel and equal in length.

4) One of the properties of a parallelogram states that when two parallel lines are cut by a transversal, alternate interior angles are congruent.

5) It can be seen from the question that △EKF≅△GKH as a result of congruence Postulate.

And the CPCTC proves,

EK ≅ GK

FK ≅ HK

CPCTC means Corresponding Parts of Congruent Triangles are Congruent.

When you have two triangles that have been proven as congruent, then each part of one triangle is equal in size and shape to the corresponding part in the other triangle.

5 0

4 years ago

I need the answer to this one

Katen [24]

Answer:

C

Step-by-step explanation:

I used a fancy calculator and that's what is gave me I use it on math tests and pass

3 0

3 years ago

We have n = 100 many random variables xi 's, where the xi 's are independent and identically distributed bernoulli random variab

Alex777 [14]

Recall that for a random variable $X$ following a Bernoulli distribution $\mathrm{Ber}(p)$ , we have the moment-generating function (MGF)

$M_X(t)=(1-p+pe^t)$

and also recall that the MGF of a sum of i.i.d. random variables is the product of the MGFs of each distribution:

$M_{X_1+\cdots+X_n}(t)=M_{X_1}(t)\times\cdots\times M_{X_n}(t)$

So for a sum of Bernoulli-distributed i.i.d. random variables $X_i$ , we have

$M_{\sum\limits_{i=1}^nX_i}(t)=\displaystyle\prod_{i=1}^n(1-p+pe^t)=(1-p+pe^t)^n$

which is the MGF of the binomial distribution $\mathcal B(n,p)$ . (Indeed, the Bernoulli distribution is identical to the binomial distribution when $n=1$ .)

8 0

3 years ago