Someone please help me with these. I forgot how to do these. Need within a hour

Find an equation of the line through the given point and perpendicular to the given line

s344n2d4d5 [400]

Answer:

Step-by-step explanation:

The equation of a straight line can be represented in the slope-intercept form, y = mx + c

Where c = intercept

For two lines to be perpendicular, the slope of one line is the negative reciprocal of the other line. The equation of the given line is

y = 2x - 2

Comparing with the slope intercept form,

Slope, m = 2

This means that the slope of the line that is perpendicular to it is -1/2

The given points are (-3, 5)

To determine c,

We will substitute m = -1/2, y = 5 and x = - 3 into the equation, y = mx + c

It becomes

5 = -1/2 × - 3 + c

5 = - 3/2 + c

c = 5 + 3/2

c = 13/2

The equation becomes

y = -x/2 + 13/2

4 0

3 years ago

A teacher places n seats to form the back row of a classroom layout. Each successive row contains two fewer seats than the prece

Alex_Xolod [135]

Answer:

The number of seat when n is odd $S_n=\frac{n^2+2n+1}{4}$

The number of seat when n is even $S_n=\frac{n^2+2n}{4}$

Step-by-step explanation:

Given that, each successive row contains two fewer seats than the preceding row.

Formula:

The sum n terms of an A.P series is

$S_n=\frac{n}{2}[2a+(n-1)d]$

$=\frac{n}{2}[a+l]$

a = first term of the series.

d= common difference.

n= number of term

l= last term

$n^{th}$ term of a A.P series is

$T_n=a+(n-1)d$

n is odd:

n,n-2,n-4,........,5,3,1

Or we can write 1,3,5,.....,n-4,n-2,n

Here a= 1 and d = second term- first term = 3-1=2

Let $t^{th}$ of the series is n.

$T_n=a+(n-1)d$

Here $T_n=n$ , n=t, a=1 and d=2

$n=1+(t-1)2$

⇒(t-1)2=n-1

$\Rightarrow t-1=\frac{n-1}{2}$

$\Rightarrow t = \frac{n-1}{2}+1$

$\Rightarrow t = \frac{n-1+2}{2}$

$\Rightarrow t = \frac{n+1}{2}$

Last term l= n,, the number of term $=\frac{ n+1}2$ , First term = 1

Total number of seat

$S_n=\frac{\frac{n+1}{2}}{2}[1+n}]$

$=\frac{{n+1}}{4}[1+n}]$

$=\frac{(1+n)^2}{4}$

$=\frac{n^2+2n+1}{4}$

n is even:

n,n-2,n-4,.......,4,2

Or we can write

2,4,.......,n-4,n-2,n

Here a= 2 and d = second term- first term = 4-2=2

Let $t^{th}$ of the series is n.

$T_n=a+(n-1)d$

Here $T_n=n$ , n=t, a=2 and d=2

$n=2+(t-1)2$

⇒(t-1)2=n-2

$\Rightarrow t-1=\frac{n-2}{2}$

$\Rightarrow t = \frac{n-2}{2}+1$

$\Rightarrow t = \frac{n-2+2}{2}$

$\Rightarrow t = \frac{n}{2}$

Last term l= n, the number of term $=\frac n2$ , First term = 2

Total number of seat

$S_n=\frac{\frac{n}{2}}{2}[2+n}]$

$=\frac{{n}}{4}[2+n}]$

$=\frac{n(2+n)}{4}$

$=\frac{n^2+2n}{4}$

4 0

2 years ago

Katie needs to buy food for her pet leopard gecko the cost of the food is $3.79 if katie uses a coupon the costs is reduced to $

LekaFEV [45]

The cost with the coupon is $0.76 less.

3.79 - 3.03

8 0

2 years ago

What is a cubic polynomial function with zeros 3,3 and -3? show your work.

Brums [2.3K]

Answer:

Cubic polynomial function with zeros 3,3 and -3 is $\mathbf{x^3-3x^2-9x+27=0}$

Step-by-step explanation:

We need to find a cubic polynomial function with zeros 3,3 and -3.

If zeros of polynomial are: 3,3,and -3

we can write:

x=3, x=3, x=-3

Or

We can write:

x-3=0, x-3=0, x+3=0

Now, we can write them as:

(x-3)(x-3)(x+3)=0

Multiplying the terms, we can find the polynomial:

$(x-3)(x-3)(x+3)=0\\(x(x-3)-3(x-3))(x+3)=0\\(x^2-3x-3x+9)(x+3)=0\\(x^2-6x+9)(x+3)=0\\x(x^2-6x+9)+3(x^2-6x+9)=0\\x^3-6x^2+9x+3x^2-18x+27=0\\x^3-6x^2+3x^2+9x-18x+27=0\\x^3-3x^2-9x+27=0$

So, cubic polynomial function with zeros 3,3 and -3 is $\mathbf{x^3-3x^2-9x+27=0}$

4 0

2 years ago

Marta hopes to have a test average of at least 90 by the end of the marking period. Her grades on the first four tests have been

Virty [35]

Answer:

Marta need to score at least 94 to receive an average of 90 or higher considering all five tests.

Explanation:

Marks received by Marta in four tests = 86, 84, 97 and 89

Let mark received in fifth test be x

We have average of five test is more than or equal to 90

So we have

$\frac{86+84+97+89+x}{5} \geq 90\\ \\ 86+84+97+89+x\geq 450\\ \\ x\geq 94$

So Marta need to score at least 94 to receive an average of 90 or higher considering all five tests.

5 0

2 years ago