Write an equation for a quadratic function in factored form with
zeros at x = -4 and x = 0 that passes through the point (-3,6).

Answer:

a=1

b=-5

c=6

Step-by-step explanation:

ax^2+bx+c=0

Is the formula used for this question.

The given function is the same as y = 1x^2 + (-5x) + 6

Compare it to the form y = ax^2+bx+c, and we have the following:

a = 1

b = -5

c = 6

Answer:

there are 6 units

hope this help

Answer:

Dilated pupils

Long periods of wakefulness

Loss of appetite

Overconfidence

Over-excitement

Paranoia

Runny nose or frequent sniffles

White powder around nostrils

Legal issues

Missing or being late to work

Financial problems

Mood swings

Irritability

Depression

Answer:

in jh ynhtym

hunt ghb JBL iggmmjuu ki ju nu j

Hello,

we are going to use the theorem of the bissectrice. (if this word is good in english)

[tex]\dfrac{KL}{LM} =\dfrac{KN}{NM} \\\\\dfrac{8}{x} =\dfrac{14}{20-x} \\\\8(20-x)=14x\\\\160-8x=14x\\\\22x=160\\\\x=\frac{160}{22} \\\\x=\frac{80}{11} \\\\\\x=7.\overline{27}[/tex]

Answer:

41

Step-by-step explanation:

Answer:

[tex]x=6[/tex]

Step-by-step explanation:

We want to solve the equation:

[tex]\displaystyle \log_6(2x-6)+\log_6x=2[/tex]

Recall the property:

[tex]\displaystyle \log_bx+\log_by=\log_b(xy)[/tex]

Hence:

[tex]\log_6(x(2x-6))=2[/tex]

Next, recall that by the definition of logarithms:

[tex]\displaystyle \log_b(a)=c\text{ if and only if } b^c=a[/tex]

Therefore:

[tex]6^2=x(2x-6)[/tex]

Solve for x. Simplify and distribute:

[tex]36=2x^2-6x[/tex]

We can divide both sides by two:

[tex]x^2-3x=18[/tex]

Subtract 18 from both sides:

[tex]x^2-3x-18=0[/tex]

Factor:

[tex](x-6)(x+3)=0[/tex]

Zero Product Property:

[tex]x-6=0\text{ or } x+3=0[/tex]

Solve for each case. Hence:

[tex]x=6\text{ or } x=-3[/tex]

Next, we must check the solutions for extraneous solutions. To do so, we can simply substitute the solutions back into the original equations and examine its validity.

Checking x = 6:

[tex]\displaystyle \begin{aligned} \log_{6}(2(6)-6)+\log_{6}6&\stackrel{?}{=} 2 \\ \\ \log_6(12-6)+(1)&\stackrel{?}{=}2 \\ \\ \log_6(6)+1&\stackrel{?}{=}2 \\ \\ 1+1=2&\stackrel{\checkmark}{=}2\end{aligned}[/tex]

Hence, x = 6 is indeed a solution.

Checking x = -3:

[tex]\displaystyle\begin{aligned} \log_6(2(-3)-6) + \underbrace{\log_6-3}_{\text{und.}} &\stackrel{?}{=} 2\\ \\ \end{aligned}[/tex]

Since the second term is undefined, x = -3 is not a solution.

Therefore, our only solution is x = 6.

Answer:

x = 6

Step-by-step explanation:

The given logarithmic equation is ,

[tex]\implies log_{6}(2x - 6) + log_{6}x = 2[/tex]

We can notice that the bases of both logarithm is same . So we can use a property of log as ,

[tex]\bf \to log_a b + log_a c = log_a {( ac)} [/tex]

So we can simplify the LHS and write it as ,

[tex]\implies log_{6} \{ x ( 2x - 6 )\} = 2 [/tex]

Now simplify out x(2x - 6 ) . We get ,

[tex]\implies log_6 ( 2x^2 - 6x ) = 2 [/tex]

Again , we know that ,

[tex]\bf \to log_a b = c , a^c = b [/tex]

Using this we have ,

[tex]\implies 2x^2 - 6x = 6^2 \\\\\implies 2x^2 - 6x -36 = 0 [/tex]

Now simplify the quadratic equation ,

[tex]\implies x^2 - 3x - 18 = 0 \\\\\implies x^2 -6x + 3x -18=0\\\\\implies x( x -6) +3( x - 6 ) = 0 \\\\\implies (x-6)(x+3) = 0 \\\\\implies x = 6 , -3 [/tex]

Since logarithms are not defined for negative numbers or zero , therefore ,

[tex]\implies 2x - 6 > 0 \\\\\implies x > 3 [/tex]

Therefore the equation is not defined at x = -3 . Hence the possible value of x is 6 .

[tex]\implies \underline{\underline{ x \quad = \quad 6 }}[/tex]

Answer:

12,-2

Step-by-step explanation:

Answer:

x = 63.9

Step-by-step explanation:

We have:

[tex]\frac{QR}{sin35}[/tex] = [tex]\frac{78}{sin90}[/tex]

=> QR × sin90 = 78 × sin35

=> QR × sin90 = 44.7

=> QR = 44.7

But, we also have: [tex]x^{2}[/tex] + [tex]QR^{2}[/tex] = [tex]78^{2}[/tex] <=> [tex]x^{2}[/tex] + [tex]44.7^{2}[/tex] = [tex]78^{2}[/tex] <=> [tex]x^{2}[/tex] = 6084 - 1998.09

=> [tex]x^{2}[/tex] = 4085.91

=> x = [tex]\sqrt{4085.91}[/tex] = 63.9211.. = 63.9

<3 Have a nice day!!

Answer:

Hello there,

The correct answer to this question would be D 1/6

Step-by-step explanation:

I had took the test and it said it was correct.

Anyways hope this helps

if the answer is correct pls mark Brainliest

thank you, have a nice day

Answer:

x = 5

arcEF = 58degrees

arcGH = 55degrees

Step-by-step explanation:

Find the diagram attached

The sum of angle on the straight line EH is 180degrees

Hence arcEF+ arcFG + arcGH = 180

10x+8 + 67 + 11x = 180

21x + 75 = 180

21x = 180 - 75

21x = 105

x = 105/21

x = 5

Since EF = 10x+8

arcEF = 10(5) + 8

arcEF = 50+8

arcEF = 58degrees

Also, arcGH = 11x

arcGH = 11(5)

arcGH = 55degrees

Answer:

B

Step-by-step explanation:

plss pa brainliest me thanks

Step-by-step explanation:

w = 34° (alternate interior angles)

x = 34° (vertically opposite angles)

y = 101°

z = 79° (corresponding angles)

Answer:

The answer would be

Step-by-step explanation:

you multiple 8 and 5 then you multiple 12 (inches) then subtract the inch that makes up 12 let me know if you need help

Answer:

Step-by-step explanation:

B/c  the bigger triangle is a copy of the smaller we can say

8[tex]\sqrt{3}[/tex] / 12  = x / 8[tex]\sqrt{3}[/tex]

then, solve for x

8[tex]\sqrt{3}[/tex](8[tex]\sqrt{3}[/tex]/12) = x

(64 *3) 12 = x

192 /12 = x

16 = x

nice,  the first answer looks good , then :P

Answer:

16

Step-by-step explanation:

Answer:

2 ( Option A )

Step-by-step explanation:

The given integral to us is ,

[tex]\longrightarrow \displaystyle \int_0^1 5x \sqrt{x}\ dx [/tex]

Here 5 is a constant so it can come out . So that,

[tex]\longrightarrow \displaystyle I = 5 \int_0^1 x \sqrt{x}\ dx [/tex]

Now we can write √x as ,

[tex]\longrightarrow I = \displaystyle 5 \int_0^1 x . x^{\frac{1}{2}} \ dx [/tex]

Simplify ,

[tex]\longrightarrow I = 5 \displaystyle \int_0^1 x^{\frac{3}{2}}\ dx [/tex]

By Power rule , the integral of x^3/2 wrt x is , 2/5x^5/2 . Therefore ,

[tex]\longrightarrow I = 5 \bigg( \dfrac{2}{5} x^{\frac{5}{2}} \bigg] ^1_0 \bigg) [/tex]

On simplifying we will get ,

[tex]\longrightarrow \underline{\underline{ I = 2 }}[/tex]

Answer:

C) 3 times the sq root of 3

Step-by-step explanation:

In a right angle triangle, the COS is definned as the ajacent side over the hypotenuse. So

COS 60 = JK/JL

COS 60 = JK/(6*sq root 3)

.5 = JK/(6*3**1/2)

JK = .5 *(6*3**1/2)

JK = 3*3**1/2

Answer:

y = - 35

Step-by-step explanation:

Expressing as a fraction , that is

[tex]\frac{y}{-7}[/tex] = 5 ( multiply both sides by - 7 to clear the fraction )

y = - 35

Answer:

1 hour

Step-by-step explanation:

hope it helps ....

Answer:

rs. 81

Step-by-step explanation:

given,

mp. = 1620

sp. = 1539

now,

discount = mp - s.p = 1620 - 1539 = 81

Answer:

€3,551.05

Step-by-step explanation:

31*29=899

899*3.65=€3,551.05

the school will get an estimate of €3,551.05

9514 1404 393

Answer:

  3.31227766..., irrational

Step-by-step explanation:

  3/20 +√10 = 0.15 + 3.16227766017...

The root of 10 is irrational, so adding that to a rational number will give an irrational sum.

The sum is about 3.31227766017..., an irrational number.

Answer:

A

Step-by-step explanation:

Answer:

to get the slope. take the change of y axis value divide by the change of x axis value

Answer:

1  4/5

Step-by-step explanation:

3/4 ÷ 5 /12

Copy dot flip

3/4 * 12/5

Rewriting

3/5 * 12/4

3/5 * 3

9/5

1 4/5

Answer:

% change in perimeter = 43. 44 %

% change in area = 16 %

Step-by-step explanation:

initial perimeter of the rectangle = 2(2p + p)

= 6p

initial area of the rectangle = 2p × p

= 2p²

therefore,

= 2p + 0.8p = 2.8 p

= p - 0.4p

= 0.6 p

= 3.4 p

= 1.68 p²

% change in perimeter

[tex] = \frac{6p - 3.4p}{6p} \times 100 \\ = \frac{2.6p}{6p} \times 100 \\ = 43.33\%[/tex]

% change in area

[tex] \frac{2 {p}^{2} - 1.68 {p}^{2} }{2 {p}^{2} } \times 100 \\ = \frac{0.32}{2} \times 100 \\ = 16\%[/tex]

Answer:

Step-by-step explanation:

Answer:

Step-by-step explanation:

Use the formulas that allow us to find the point that divides the segment into a ratio which is

[tex]x=\frac{bx_1+ax_2}{a+b}[/tex]  and  [tex]y=\frac{by_1+ay_2}{a+b}[/tex] where a is 1 (comes from the ratio) and b is 3 (comes from the ratio as well). Filling in:

[tex]x=\frac{3(9)+1(6)}{1+3}=\frac{27+6}{4}=\frac{33}{4}[/tex] and then y:

[tex]y=\frac{3(-1)+1(-10)}{1+3}=\frac{-3-10}{4}=-\frac{13}{4}[/tex] so the coordinates in question are

[tex](\frac{33}{4},-\frac{13}{4})[/tex]

Answer:

Step-by-step explanation:

the answer would be (7,-7)

Answer:

d=3

Step-by-step explanation:

13d=39

d=39/13

d=3

Hello!

6d + 7d = 39 <=>

<=> 13d = 39 <=>

<=> d = 39 : 13 <=>

<=> d = 3 => 6 × 3 + 7 × 3 = 39

Good luck! :)