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3 years ago

15

Given a quadratic function, y = ax2 + bx + c, what happens to the graph when "a" is positive? What happens to the graph when "a"

is negative?

1 answer:

Elanso [62]3 years ago

4 0

Answer:

when a is positive, the graph is a smiling graph. when a is negative the curve is a sad graph. thats how i remember it

You might be interested in

5/6*2/3<br> 9/10*5/18<br> 4/5*3/4<br> 2/3*5/1

zalisa [80]

To multiply fractions, you multiply numerator with numerator and denominator with denominator

In other words: a/b*c/d=(a*c)/(b*d)

So 1st one is (5*2)/(6*3) which is 10/18 or 5/9 simplified

2nd one is (9*5)/(10*18) which is 45/180 or 1/4 simplified

3rd one is (4*3)/(5*4) which is 12/20 or 3/5 simplified

4th one is (2*5)/(3*1) which is 10/3 or 3 1/3 if you want a mixed number

Hope this helped!

5 0

3 years ago

Read 2 more answers

Jim is enclosing a rectangular garden with 170 feet of fencing. The length of the garden is 10 feet more than twice it’s width,

Darina [25.2K]

Answer:

$1,500\ ft^2$

Step-by-step explanation:

Let

L ----> the length of the rectangular garden in feet

w ---> the width of the rectangular garden in feet

step 1

Find the width

we know that

The perimeter of the rectangular garden is

$P=2(L+W)$

$P=170\ ft$

so

$170=2(L+W)$

Simplify

$85=(L+W)$ ----> equation A

$L=2W+10$ ----> equation B

substitute equation B in equation A and solve for W

$85=(2W+10+W)$

$85-10=3W$

$3W=75$

$W=25\ ft$

Find the value of L

$L=2(25)+10=60\ ft$

step 2

Find the area

we know that

The area of the rectangular garden is

$A=(LW)$

substitute the values

$A=(60)(25)=1,500\ ft^2$

6 0

3 years ago

Factor the following polynomial: x2−5x+6

inessss [21]

Answer:

(x-2)(x-3)

Step-by-step explanation:

7 0

3 years ago

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Steven wishes to save for his retirement by depositing $2,000 at the beginning of each year for thirty years. Exactly one year a

Ad libitum [116K]

Step-by-step explanation:

i = interest 3% for 30 years

This is a simple dynamical system for whom the the solutions are given as

$S=R[\frac{(i+1)^n-1}{i}](i+1)$

putting values we get

S=2000[\frac{(1.03)^{30}-1}{0.03}](1.03)

= $98005.35

withdrawal of money takes place from one year after last payment

To determine the result we use the present value formula of an annuity date

$P = R\frac{1-(1+i)^{-n}}{i}{i+1}$

we need to calculate R so putting the values and solving for R we get

R= $6542.2356

8 0

3 years ago

Factor the expression -10x^2-41x-21

BaLLatris [955]

$-10x^2-41x-21= -(10x^2+41x+21) =-(10x^2+35x+6x+21)=\\\\=- [(10x^2+35x)+(6x+21 )]=- [5x(2x +7 )+3(2x+7 )]=-(2x+7)(5x+3)$

5 0

2 years ago

Read 2 more answers