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3 years ago

Help with some algebra please?

1 answer:

7 0

Vertex form is f(x) = a(x - b)^2 + c where the vertex is (b, c) and a is some constant.

So for the first part we have f(x) = a(x + 6)^2 - 1

for root -9:- a( -9+6)^2 - 1 = 0
9a - 1 = 0 , a = 1/9

The other 3 are derived in the same way.
so required equation is f(x) = (1/9)(x + 6)^2 - 1

You might be interested in

Find the slope of the line y = 5x + 4.

avanturin [10]

Answer:

x=1/5y+−4/5

Step-by-step explanation:

Let's solve for x.

y=5x+4

Step 1: Flip the equation.

5x+4=y

Step 2: Add -4 to both sides.

5x+4+−4=y+−4/5x=y−4

Step 3: Divide both sides by 5.

5x/5=y−4/5x=1'5y+−4'5

Answer:

x=1/5y+−4/5

( I hope this was helpful) >;D

6 0

3 years ago

What is the product of 5.06 and 2.1

Oksana_A [137]

Answer:

10.626

Step-by-step explanation:

There's multiple ways to go about it. See how

5.06 * 2.1 = (5 + 0.06)(2 + 0.1) = 5*2 + 0.06*2 + 5*0.1 + 0.06*0.1 =

= 10 + 0.12 + 0.5 + 0.006 = 10.626

Please notice how breaking up the product into smaller parts has let us make all the computations without too much hassle.

7 0

2 years ago

Read 2 more answers

Find the area of the following<br> rhombus:<br> 6 cm<br> 10 cm<br> 8 cm

alexira [117]

Answer:

480

Step-by-step explanation:

10*6=60

60*8=480

If my math is correct. I am sorry if I am wrong

8 0

3 years ago

ASAP ASAP ASAP ASAP MATH MATH MATH homework

Brut [27]

Answer:

$20

Step-by-step explanation:

Simple, if he sold 16 pretzels, and each one costs $1.25, then multiply the cost by how many:

16X1.25=20

So he got $20

I hope this helps!

5 0

3 years ago

In a shipment of 56 vials, only 13 do not have hairline cracks. If you randomly select 3 vials from the shipment, in how many wa

Elden [556K]

Answer: 27434

Step-by-step explanation:

Given : Total number of vials = 56

Number of vials that do not have hairline cracks = 13

Then, Number of vials that have hairline cracks =56-13=43

Since , order of selection is not mattering here , so we combinations to find the number of ways.

The number of combinations of m thing r things at a time is given by :-

$^nC_r=\dfrac{n!}{r!(n-r)!}$

Now, the number of ways to select at least one out of 3 vials have a hairline crack will be :-

$^{13}C_2\cdot ^{43}C_{1}+^{13}C_{1}\cdot ^{43}C_{2}+^{13}C_0\cdot ^{43}C_{3}\\\\=\dfrac{13!}{2!(13-2)!}\cdot\dfrac{43!}{1!(42)!}+\dfrac{13!}{1!(12)!}\cdot\dfrac{43!}{2!(41)!}+\dfrac{13!}{0!(13)!}\cdot\dfrac{43!}{3!(40)!}\\\\=\dfrac{13\times12\times11!}{2\times11!}\cdot (43)+(13)\cdot\dfrac{43\times42\times41!}{2\times41!}+(1)\dfrac{43\times42\times41\times40!}{6\times40!}\\\\=3354+11739+12341=27434$

Hence, the required number of ways =27434

5 0

3 years ago