can someone help me out

Answer:x²-3 x+4/ 2x-3 units

Step-by-step explanation:

Area = Length * Width sq. units

x^2 - 3x + 4 = Length * 2x - 3

=>Length = x^2- 3 x + 4/ 2 x − 3 units

Answer:

h and the rest are the best for me 7-Eleven you have to go for your

Step-by-step explanation:

happy new years and hope you have not forgotten you and have been busy happy weekend to your the as is have weekend lot here well the there is you idom not you big very for hard while and there can't a help night get be to good in the world of the world and the world is a doctor that is a good friend and is not a good person .

Answer:

Step-by-step explanation:

I'm going to use Physics here for this concept of vectors. Here are some stipulations I have set for the problem (aka rules I set and then followed throughout the problem):

** I am counting the 50 m as 2 significant digits even though it is only 1, and I am counting 200 as 3 significant digits even though it is only 1. 1 sig dig doesn't really give us enough accuracy, in my opinion.

** A bearing of .25 degrees is measured from the North and goes clockwise; that means that measured from the x axis, the angle is 89.75 degrees. This is the angle that is used in place of the bearing of .25 degrees.

** Due east has an angle measure of 0 degrees

Now let's begin.

We need to find the x and y components of both of these vectors. I am going to call the first vector A and the second B, while the resultant vector will be C. Starting with the x components of A and B:

[tex]A_x=50cos(89.75)[/tex] so

[tex]A_x=.22[/tex]

[tex]B_x=200cos(0)[/tex] so

[tex]B_x=200[/tex] and we need to add those results together. Due to the rules for adding significant digits properly, the answer is

[tex]C_x=200[/tex] (and remember I am counting that as 3 sig fig's even though it's only 1).

Now for the y components:

[tex]A_y=50sin(89.75)[/tex] so

[tex]A_y=50[/tex] (which I'm counting as 2 sig fig's)

[tex]B_y=200sin(0)[/tex] so

[tex]B_y=0[/tex] and we need to add those results together.

[tex]C_y=50[/tex]

Now for the resultant magnitude:

[tex]C_{mag}=\sqrt{(200)^2+(50)^2}[/tex]  and that gives us a final magnitude of

[tex]C_{mag}=206[/tex] m

Now for the angle:

Since both the x and y components of the resultant vector are in quadrant 1, we don't need to add anything to the angle to get it right, so

[tex]tan^{-1}(\frac{50}{200})=14[/tex]

The girl is 206 meters from her starting point at an angle of 14 degrees

Answer:

Its that squiglly one on top

Step-by-step explanation:

cuh cuh cuh cuh cuhc

Answer:

Area of pizza slice = 87 inches² (Approx.)

Step-by-step explanation:

Given:

Angle cut of pizza = 100°

Diameter of pizza = 20 inches

Find:

Area of pizza slice

Computation:

Area of sector = [Ф / 360][π][r²]

Area of pizza slice = [100/360][3.14][(20/2)²]

Area of pizza slice = [100/360][3.14][(10)²]

Area of pizza slice = [100/360][3.14][100]

Area of pizza slice = [0.2778][3.14][100]

Area of pizza slice = 87.2292

Area of pizza slice = 87 inches² (Approx.)

Answer:

There is not enough evidence to support the claim that drug is effective at 1%

Step-by-step explanation:

For this we use the pored test as we have dependent test;

Difference :

10,5,-5,3,-3,4,2,-3,0,10,8,-2,3,-2,5,5,4,0

Hypothesis :

H0 : μd = 0

H0 : μd ≠ 0

The mean of d, dbar = Σd / n = 44 / 18 = 2.44

The standard deviation, Sd = 4.45 (using calculator)

The test statistic, T : dbar / (Sd/√n)

T = 2.44 / (4.45/√18)

T = 2.44 / 1.0488750

T = 2.326

Degree of freedom, df = n - 1 ; 18 - 1 = 17

The Pvalue = 0.0326

α = 0.01

Since Pvalue < α ; we fail to reject H0

Answer:

C. -7

Step-by-step explanation:

See the picture for steps :)

Answer:

Step-by-step explanation:

Since T is in Quadrant II, sinT > 0 and cosT < 0.

cotT = -5/12

Answer:

jsjzbdjwbxjdbwisjsobs9fbe

Step-by-step explanation:

hfkdowwhdbhdvsjsh2ivdusvqvduzvqidv

Answer:

Yeeeeaaahhhh its correct.

Answer:

yes, your answers are correct

The perimeter = sum of all sides

= 120 + 80 + 50

= 250

So 250 - 3

247

Left space for gate

Now cost of fencing = Rs 20/per meter

= 247 × 20

= Rs 4,940

Now the area of the triangular park can be found using heron's formula

S = (a+b+c)/2

S = (120+80+50)/2

S = 250/2

S = 125

Now

Herons formula = √s(s-a)(s-b)(s-c)

√125(125-120)(125-80)(120-50)

√125(5)(45)(70)

√5×5×5×5×5×3×3×5×14

After Making pairs

5×5×5×3√14

375√14

Therefore 375√14m is the area of the triangular park

Must click thanks and mark brainliest

$\sf\underline\bold{Answer:}$

$\space$

$\sf\underline\bold{Step-by-Step:}$

$\space$

$\sf\bold{Given(In \:the\:Q):}$

$\space$

$\sf\bold{To \: find:}$

$\space$

$\sf\small{☆Area\:to\:be\:planted=Area \: of \: ∆ABC}$

$\space$

$\sf\underline\bold{Calculating\:area\:of\:∆ABC:}$

$\space$

Use heron's formula to find the area of the triangle.

$\space$

$\mapsto$ $\sf\small{Heron's\:formula=}$

[tex]\sf\sqrt{s(s-a)(s-b)(s-c)}[/tex]

$\space$

$\sf\bold{Now,find\:semi\:parameter:-}$

$\sf\small{Perimeter\:of\:the\:∆=120+80+50=250}$

$\sf\small{Semi-Perimeter:}$ $\sf\dfrac{250}{2}$ $\sf\small{=125m}$

$\space$

$\sf\small{Substitute \: the\:values\:in\:heron's\:formula:}$

$\sf{Area\:of\:the\:∆:-}$

$\mapsto$ [tex]\sf\sqrt{125(125-120)(125-80)(125-50)}[/tex]

$\space$

$\mapsto$ $\sf\sqrt{125\times(5)\times(45)\times(75)}$

$\space$

$\mapsto$ $\sf\small\sqrt{2109375}$ $\sf\small{=375}$ $\sf\small\sqrt{15}$

$\space$

$\longmapsto$ $\sf\underline\bold\purple{1452.56m^2}$

______________________________

$\sf\underline\bold{Now,find\:the\:cost\:of\:fencing:}$

$\sf{Cost\:of\:fencing-}$

$\space$

$\sf\underline{Hence,the\:gardener\:has\:to\:fence:}$

$\space$

So,total cost of fencing at the rate of Rs.20 per m:-

___________________________________

Answer:

hello,

answer C

Step-by-step explanation:

area of the rectangle=2b*b=2b²

area of the half cercle= [tex]\\\dfrac{\pi*(\frac{c}{2})^2}{2} =\dfrac{\pi*c^2}{8} \\[/tex]

Area in blue= 2b²-πc²/8

Answer:

-7*6^n

Step-by-step explanation:

The sequence is -7*6^n.

A few examples of this are the following,

ex.)

-7*6^0 = -7

-7*6^1 = -42

-7*6^2 = -252

Answer:

[tex]{26}[/tex]

Step-by-step explanation:

[tex] \sqrt{(0 - ( - 5) ){}^{2} + (0 - 1) {}^{2} } [/tex]

[tex] = \sqrt{ {(5)}^{2} + {( - 1)}^{2} } [/tex]

[tex] = \sqrt{25 + 1} [/tex]

[tex] = \sqrt{26} [/tex]

Answered by GAUTHMATH

Answer:

26

Step-by-step explanation:

(-5, 1) (0, 0)

sqrt(5^2 + (-1^2))

sqrt(25 + 1)

sqrt(26)

So, 26 goes under the radical.

Answer:

The right answer is:

(a) 0.4866

(b) 10.3972

Step-by-step explanation:

Let X be the number of days,

X is [tex]exp(\lambda)[/tex],

⇒ [tex]\lambda =\frac{1}{15}[/tex]

now,

⇒ [tex]F_x(x) = P(X \leq x)[/tex]

              [tex]=1-e^{-\lambda x}[/tex]

              [tex]=1-e^{-\frac{x}{15} }[/tex]

or,

[tex]x>0[/tex]

(a)

⇒ [tex]P(X<10)=F_x(10)[/tex]

                      [tex]=1-e^{\frac{10}{15} }[/tex]

                      [tex]=1-e^{\frac{2}{3} }[/tex]

                      [tex]=0.4866[/tex]

(b)

If the medium be m, then

[tex]P(X=m) = \frac{1}{2}[/tex]

                 [tex]=2^{-1}[/tex]

⇒[tex]1-e^{-\frac{m}{15} } = \frac{1}{2}[/tex]

        [tex]e^{-\frac{m}{15} } = 2^{-1}[/tex]

             [tex]m= 15\ ln2[/tex]

                 [tex]=10.3972[/tex]  

Answer:

filtration system is not working

filtration

Answer:

y = 6x

Step-by-step explanation:

The equation representing a proportional relationship is

y = kx ← k is the constant pf proportionality

Here k = 6 , then

y = 6x ← equation of proportion

Answer:

She is not correct, she would need to have something before the y to know the answer

Step-by-step explanation:

hope this helps :)

Answer:

a = m/c

~~~~~~~~~~~~~~~~~~~~~~~

m=ca

divide both sides by c

you get a = m/c

Step-by-step explanation:

Answer:

a = m/c

Step-by-step explanation:

m = ca

divide both sides by c

m/c = ca/c

m/c = a

:.a = m/c

Answer:

x³-2x²-4x+8

Step-by-step explanation:

a1=x+2

a2=x²-4⇒q=a2/a1=x-2

⇒a3=q.a2=(x-2).(x²-4)=x³-2x²-4x+8

Answer:

keep the pic clearly

Step-by-step explanation:

keep the pic clearly

Answer:

I got that AE^2 = 8EC^2

Step-by-step explanation:

Here's what I did. I probably did this wrong, but this might give you an idea on how to solve it.

First, create equations:

AB = BC = AC = DE = 2EC

BD= BC = EC

Then, I used pythagorean theorem (since AD is perpendicular to BC): AD^2 + DE^2 = AE^2

Substitute: (2EC)^2 + (2EC)^2 = AE^2

Simplify: 4EC^2 + 4EC^2 = AE^2

AE^2 = 8CE^2

Answer:

(I don't have the complement sign of sets, so I've used "©" to denote compliment)

L.H.S.

(B-A)∪C

= (B∩A©)∪C

= (B∪C)∩(A©∪C)

= (B∪C)∩(A-C)©

= (B∪C)-(A-C) = R.H.S.

(Proved)

Answer:

a)-2x(x+4x²)+3(x²+2x)

-2x²-8x³+3x²+6x

-2x²+3x²+6x-8x³

x²-8x³+6x

in descending order

-8x³+x²+6x

b)(4x-3)(4x+3)

4x(4x+3)-3(4x+3)

16x²+12x-12x-9

16x²-9

I hope this helps and sorry if it's wrong

To find the time in which all the bells ring together, we need to find the LCM of 36,40,48.

Prime factorization of 36=2×2×3×3

Prime factorization of 40=2×2×2×5

Prime factorization of 48=2×2×2×2×3

Hence, LCM of 36,40,48=2×2×2×2×3×3×5=720 seconds.

720seconds=  

60

720

​

minutes=12minutes

Hence, all the bells will ring together after 12 mi

Answer:

9pm or 12 hours from 9am.

Answer:

5

Step-by-step explanation:

Since this is a right triangle, we can use the Pythagorean theorem

a^2+b^2 = c^2  where a and b are the legs and c is the hypotenuse

a^2 + 10^2 = (5 sqrt(5))^2

a^2 +100 = 25(5)

a^2 +100 = 125

a^2 = 125-100

a^2 = 25

Taking the square root of each side

sqrt(a^2) = sqrt(25)

a = 5

Answer:

7i/sqrt(2)

Step-by-step explanation:

so, if I understand correctly, the equation is

(22-16)(2x² + 49) = 0

6(2x² + 49) = 0

that automatically means

2x² + 49 = 0

2x² = -49

x² = -49/2

x = sqrt(-49/2)

a square root of a negative number has no real solution. the solution is in the world of complex and imaginary numbers, which bases on the sqrt(-1) = i.

so, we get

x = sqrt(49/2 × -1) = (7/sqrt(2)) × I = 7i/sqrt(2)

Answer:  Choice A

y=2x+3; y=-1/3x+3

==================================================

Explanation:

The left portion of the blue curve is y = 2x+3 but it is only graphed when x < 0 (we could argue that [tex]x \le 0[/tex] but I'll set that aside for the other portion).

The right portion is the line y = -1/3x + 3 and it's only graphed when [tex]x \ge 0[/tex]

So we could have this piecewise function

[tex]f(x) = \begin{cases}2x+3 \ \text{ if } x < 0\\-\frac{1}{3}x+3 \ \text{ if } x \ge 0\\\end{cases}[/tex]

Or we could easily swap the "or equal to" portion to move to the first part instead like this

[tex]f(x) = \begin{cases}2x+3 \ \text{ if } x \le 0\\-\frac{1}{3}x+3 \ \text{ if } x > 0\\\end{cases}[/tex]

Either way, we're involving the equations mentioned in choice A