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2 years ago

Find the range of the graphed function?

Mathematics

2 answers:

Andrei [34K]2 years ago

7 0

For finding range pay attention to y axis max of function is 5 and its min is -9
B is true

frez [133]2 years ago

7 0

B is the correct answer

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Can anyone help me please I’ll mark as brainliest.

sergejj [24]

Answer:

First tables blanks are 160, 195, 230, 300

The rate of change would be 35/1 cos the cost is going up by 35 and the months are going up by one

The second tables first row is 1 2 3 4 5, the second row is 5 7 9 11 13

5 0

2 years ago

A square has a perimeter of 12x 52 units. which expression represents the side length of the square in units? x 4 x 40 3x 13 3x

d1i1m1o1n [39]

The expression which represents the side length of the square which has perimeter of 12x 52 units is 3x+12 units.

<h3>What is the perimeter of a square?</h3>

The measure of the boundary of the sides of a square is called its perimeter. The perimeter of a square is 4 times the side of the square.

$P=4a$

Here, (a) is the side of the square.

A square has a perimeter of 12x +52 units. The perimeter of a square is 4 times the side of the square. Thus, the side of this square is,

$P=4a\\12x +52=4a\\a=\dfrac{12x +52}{4}\\a=\dfrac{4(3x +12)}{4}\\a=3x+12$

Thus, the expression which represents the side length of the square which has perimeter of 12x 52 units is 3x+12 units.

Learn more about the perimeter of the square here;

brainly.com/question/25092270

7 0

1 year ago

Find the sum of all natural numbers between 25 and 210 which are either divisible by 3 or divisible by 4? Please let me know I

Ray Of Light [21]

Let S be the sum of the integers 25-210:

S = 25 + 26 + 27 + … + 208 + 209 + 210

Let S₃, S₄, and S₁₂ denote the sums of the integers in S that are multiples of 3, 4, or 12, respectively. We'll also count how many terms each sum involves; it'll be useful later.

S₃ = 27 + 30 + 33 + … + 204 + 207 + 210 … … … (62 terms)

S₃ = 3 (9 + 10 + 11 + … + 68 + 69 + 70)

S₄ = 28 + 32 + 36 + … + 200 + 204 + 208 … … … (46 terms)

S₄ = 4 (7 + 8 + 9 + … + 50 + 51 + 52)

S₁₂ = 36 + 48 + 60 + … + 180 + 192 + 204 … … … (15 terms)

S₁₂ = 12 (3 + 4 + 5 + … + 15 + 16 + 17)

Let's look at S₃ :

S₃ = 3 (9 + 10 + 11 + … + 68 + 69 + 70)

By reversing the order of the sum, we get

S₃* = 3 (70 + 69 + 68 + … + 11 + 10 + 9)

Of course S₃ = S₃*, I'm just calling it something else temporarily. Notice that every term in the same position of either sum adds up the same number.

9 + 70 = 79

10 + 69 = 79

11 + 68 = 79

and so on. Then

S₃ + S₃* = 3 (79 + 79 + 79 + … + 79 + 79 + 79)

2S₃ = 3 × 62 × 79 ==> S₃ = 7,347

We can compute the other two sums in the same way.

S₄ = 4 (7 + 8 + 9 + … + 50 + 51 + 52)

S₄* = 4 (52 + 51 + 50 + … + 9 + 8 + 7)

==> 2S₄ = 4 × 46 × 59 ==> S₄ = 5,428

S₁₂ = 12 (3 + 4 + 5 + … + 15 + 16 + 17)

S₁₂* = 12 (17 + 16 +15 + … + 5 + 4 + 3)

==> 2S₁₂ = 12 × 15 × 20 ==> S₁₂ = 1,800

Then the sum you want is

S₃ + S₄ - S₁₂ = 10,975

We subtract S₁₂ because each of its terms is counted twice (once in S₃ and again in S₄).

6 0

2 years ago

You guy have any song requests?

scoundrel [369]

Answer:

Never want to give you up

Step-by-step explanation:

Brainliest? pls

8 0

3 years ago

Read 2 more answers

Find the equation of a line passing through (-4,-3) and perpendicular to 3 x + 2 y = 14.

Reil [10]

keeping in mind that perpendicular lines have negative reciprocal slopes, let's check for the slope of the equation above

$3x+2y=14\implies 2y=-3x+14\implies y=\cfrac{-3x+14}{2}\implies y = \cfrac{-3x}{2}+\cfrac{14}{2} \\\\\\ y=\stackrel{\stackrel{m}{\downarrow }}{-\cfrac{3}{2}}x+7\qquad \impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}$

so therefore

$\stackrel{~\hspace{5em}\textit{perpendicular lines have \underline{negative reciprocal} slopes}~\hspace{5em}} {\stackrel{slope}{\cfrac{-3}{2}} ~\hfill \stackrel{reciprocal}{\cfrac{2}{-3}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{2}{-3}\implies \cfrac{2}{3}}}$

so we're really looking for the equation of a line whose slope is 2/3 and passes through (-4 , -3)

$(\stackrel{x_1}{-4}~,~\stackrel{y_1}{-3})\qquad \qquad \stackrel{slope}{m}\implies \cfrac{2}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-3)}=\stackrel{m}{\cfrac{2}{3}}(x-\stackrel{x_1}{(-4)}) \implies y+3=\cfrac{2}{3}(x+4) \\\\\\ y+3=\cfrac{2}{3}x+\cfrac{8}{3}\implies y=\cfrac{2}{3}x+\cfrac{8}{3}-3\implies y=\cfrac{2}{3}x-\cfrac{1}{3}$

3 0

2 years ago