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2 years ago

Help help!!!!!!!!!!!!!!!!!!!!!!!!!!!

Mathematics

1 answer:

LenaWriter [7]2 years ago

6 0

Answer:

The first option, the mayor will lose by a large percentage.

Step-by-step explanation:

If you look at the numbers, You can see that the two best answers which are excellent and satisfactory are low, and in total equal 28.

Being compared to the lower ratings which are needs improvement and unsatisfactory are high, and in total equal 52.

Seeing this, this shows that the mayor will lose the election the with more than half of the people voting for someone else.

All the other options excluding the first one, say that he will win by a close call, a big percent, or a close call again.

which just leaves the first option saying that he will lose by large percent, which is true.

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Given f(x) = 3x − 4 and g(x) = f(2x), which table represents g(x)?

love history [14]

Answer:

answer is d

Step-by-step explanation:

from question

g(x) = 2(3x) - 4 = 6x - 4

then put x = 1,2,3

so g(x) = 2,8,14

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2 years ago

$54,000 at 7% for 3 years

solniwko [45]

The answer to this problem is 11340
Multiply 54,000 by 7% (7/100) by 3
remember i=prt ( interest= principal, rate, time

5 0

3 years ago

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In a circle of radius 10 cm, a sector has an area of 40 (pi) sq. cm. What is the degree measure of the arc of the sector?

Rashid [163]

Answer:

144°

Step-by-step explanation:

First, find the area of the circle, with the formula A = $\pi$ r²

Plug in 10 as the radius, and solve

A = $\pi$ r²

A = $\pi$ (10²)

A = 100 $\pi$

Using this, create a proportion that relates the area of the sector to the degree measure of the arc.

Let x represent the degree measure of the arc of the sector:

$\frac{40\pi }{100\pi }$ = $\frac{x}{360}$

Cross multiply and solve for x:

100 $\pi$ x = 14400 $\pi$

x = 144

So, the degree measure of the sector arc is 144°

6 0

3 years ago

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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

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

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

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

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

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2 years ago

An art class is making paperweights out of plaster. They use a mold in the shape of a rectangular prism. (2 1/4m, 2 1/2m, and 1

djyliett [7]

V=w•h•l
w=2 1/2=5/2=2.5 m
h=1 3/4=7/4=1.75 m
l= 2 1/4=9/4=2.25 m
V= 2.5 •1.75 •2.25
V=9.84 m^3

3 0

2 years ago

Read 2 more answers