All categories

2 years ago

PLEASE ANSWER 100 PTS AND WILL MARK BRAINLIEST.

Mathematics

2 answers:

Alexxandr [17]2 years ago

6 0

The answer is 1! Hope that helps a little-

timofeeve [1]2 years ago

3 0

Answer:

Step-by-step explanation:

We have the graph of f(x).

In order to find f(0), we simply need to find the y-coordinate of the coordinate when x is 0.

From the graph, we can see that there is a point at (0,1).

In other words, when x is 0, f(x) is 1.

Therefore:

$f(0)=1$

You might be interested in

Find the area of an equilateral triangle (regular 3 Gon) with the given measurement 9 in perimeter a=SQ.in

blsea [12.9K]

Answer:

3.897 $in^{2}$

Step-by-step explanation:

equilateral triangles are also equiangular, meaning the have equal angles.

Triangle sum theory says that angles of a triangle add up to 180.

That means each angle is 60.

A = bh/2

You need the (h). The base of is 3. Perimeter = 9, so each side is 3

Draw a perpendicular line for the height. The line cuts the base in half (1.5)

Using trigonometry you can find the height.

tan 60° = h/1.5

h = height, 1.5 is half of 3, 60° is the base angle.

multiply each side by 1.5

1.5(tan 60°) = h

h=2.598

then substitute h into formula

A= <u>(2.598)(3) </u>

A = 3.897 rounded

4 0

2 years ago

Use 3.14 for r. Round your answers to

galben [10]

Answer:

I think its A........ I don't really know.

Step-by-step explanation:

5 x 2 = 10

3.14 x 10 = 31.4

3 0

2 years ago

Greg is trying to solve a puzzle where he has to figure out two numbers, x and y. Three less than two-third of x is greater than

Fittoniya [83]

Inequation 1:

$\frac{2}{3}x-3 \geq y$

to plot the pairs (x, y) for which the inequation holds, draw the line $y=\frac{2}{3}x-3$

then pick a point in either side of the line. If that point is a solution of the inequation, than color that region of the line, if that point is not a solution, then color the other part of the line.

we do the same for the second inequation. Then the solution, is the region of the x-y axes colored in both cases.

inequation 2:

$y+ \frac{2}{3}x\ \textless \ 4$

$y\ \textless \ - \frac{2}{3} x+ 4
$

draw the lines

i) $y=\frac{2}{3}x-3$ use points (0, -3), (3, -1)

ii) $y=- \frac{2}{3} x+ 4$ use points ( 0, 4), (3, 2)

let's use the point P(3, 3) to see what region of the lines need to be coloured:

$\frac{2}{3}x-3 \geq y$ ;
$\frac{2}{3}(3)-3 \geq 3$
$2-3 \geq 3$ , not true so we color the region not containing this point

$y+ \frac{2}{3}x\ \textless \ 4$
$(3)+ \frac{2}{3}(3)\ \textless \ 4$
$3+ 5\ \textless \ 4$ not true, so we color the region not containing the point (3, 3)

The graph representing the system of inequalities is the region colored both red and blue, with the blue line not dashed, and the red line dashed.