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4 years ago

14

...............................

2 answers:

tatiyna4 years ago

8 0

I believe it would be y = 3x

Hope this helps!

erma4kov [3.2K]4 years ago

7 0

So if you know how to find slope this is a pretty easy question to solve.

Sooo,
to solve you would 2 pairs of #'s off the graph and subtract theM
Like so;
I will use the point (2,6) and (3,9).
To subtract you are going to flip the values and subtract.
9-6=3
3-2=1
This was going to represented by a fraction but since 3/1=3 the slope is just going to be 3.

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ioda

Answer:

$x = \sqrt{\frac{7}{2} }$

Step-by-step explanation:

<u><em>Explanation</em></u>

From given graph

AC = √7

Given angle ∝ = 45°

$cos\alpha = \frac{adjacentside}{Opposite side}$

$cos45 = \frac{x}{\sqrt{7} }$

$\frac{1}{\sqrt{2} } = \frac{x}{\sqrt{7} }$

$x = \frac{\sqrt{7} }{\sqrt{2} }$

$x = \sqrt{\frac{7}{2} }$

x = 1.8708

7 0

3 years ago

Is 2.67034165508… rational or unrational?

ivolga24 [154]

Answer - Yes it’s rational

4 0

3 years ago

4. Brett is fencing in a circular patio. The radius of the patio is 70 feet. How many linear feet of

skad [1K]

Answer:

The number of feet of fencing that he needs will be equal to the perimeter of the circular patio.

We know that the perimeter of a circle of radius R is equal to:

P = 2*pi*R

where pi = 3.14

Then if the radius of the patio is 70ft, the perimeter of the patio will be:

P = 2*3.14*70ft = 439.6 ft

Then he needs 439.6 ft of fencing to enclose the patio.

4 0

3 years ago

Two points sit on a line with the following coordinates p(2,2) and Q(3,7) which is the distance from p to q

Alja [10]

ANSWER

$\sqrt{26}$

EXPLANATION

Given the coordinates p(2,2) and Q(3,7),

we can use the distance formula to find the distance between P and Q.

The distance formula is given by,

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

We substitute the coordinates to get,

$d=\sqrt{(3-2)^2+(7-2)^2}$

$d=\sqrt{(1)^2+(5)^2}$

$d=\sqrt{1+25}$

$d=\sqrt{26}$

8 0

3 years ago

Find the missing side . round the the nearest tenth

lys-0071 [83]

$S\frac{O}{H} C\frac{A}{H} T\frac{O}{A}$

$Sin=\frac{Opposite}{Hypotenuse} Cos=\frac{Adjacent}{Hypotenuse} Tan=\frac{Opposite}{Adjacent}$

7) You have the opposite. You need the adjacent.

$Tan=\frac{Opposite}{Adjacent}$

$Tan57=\frac{12}{x}$

$x=\frac{12}{Tan57}$

x = 7.79289... =

<h2>7.8</h2><h2 />

8) You have the hypotenuse. You need the Opposite.

$Sin=\frac{x}{Hypotenuse}$

$Sin37=\frac{x}{13}$

x = sin37 × 13

x = 7.8235... =

<h2>7.8</h2><h2 />

9) You have the hypotenuse. You need the opposite.

$Sin=\frac{x}{Hypotenuse}$

$Sin59=\frac{x}{11}$

x = sin59 × 11

x = 9.4288.... =

<h2>9.4</h2><h2 />

10) You have the hypotenuse. You need the opposite.

$Sin=\frac{x}{Hypotenuse}$

$Sin53=\frac{x}{11}$

x = sin53 × 11

x = 8.7459.... =

<h2>8.7</h2>

7 0

3 years ago