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4 years ago

Which answer is the graph of Y<1-3x?

Mathematics

2 answers:

Romashka [77]4 years ago

7 0

Answer:

The question is unclear

Step-by-step explanation:

Mekhanik [1.2K]4 years ago

3 0

Answer:

Step-by-step explanation:

y < 1%2F3x + 1

First draw the boundary line y = 1%2F3%29x + 1

Draw it dotted because the inequality is "less than",

not "less than or equal to".

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Helpppppppppppppppppppp pleaseeeeeeeeeee

Umnica [9.8K]

Answer:

FG = 16

Step-by-step explanation:

if df bisects <edg then triangles DFE and DFG are similar AAA

since DF is the same in both triangles triangles DFE and DFG are congruent

therefore FE = FG

n+5 = 2n-6

subtract n from each side

n-n+5 = 2n-6-n

5 = n-6

add 6 to each side

5+6 = n

n=11

FG = 2n-6

=2*11 -6

= 22-6

= 16

6 0

3 years ago

the cost of four scarves and six hats is $52.00.the cost of two hats is $1 more then the cost of 1 scarf.how much does 1 scarf c

Elden [556K]

Please elaborate more on your question so I can help you

7 0

3 years ago

Can anyone please help me out with this?

nika2105 [10]

It works well to write the given angles on the diagram, then make use of the relationships for right angles and triangles.

8 0

3 years ago

Plz help asap this table represent a quadratic function with a vertex a (1,2) what is the average rate of change for the interva

Advocard [28]

Answer:

Step-by-step explanation:

Given that (1, 2) is the vertex of the function represented by the table of values above, the rate of change for the interval from x = 5 to x = 6.

f(5) = 18

f(6) = ?

=>Find f(6) using the vertex form function, f(x) = a(x - h)² + k

Where, h and k are the given vertex of the function = (1, 2).

h = 1, k = 2

Thus,

f(x) = a(x - 1)² + 2

Find the value of a by using any of the points given in the table.

Using (3, 6), we have the following,

6 = a(3 - 1)² + 2

6 = a(2)² + 2

6 = 4a + 2

Subtract 2 from both sides

6 - 2 = 4a

4 = 4a

Divide both sides by 4

1 = a

a = 1

Let's find f(6) using f(x) = a(x - 1)² + 2

Plug the value of a and x

f(6) = 1(6 - 1)² + 2

f(6) = 25 + 2

f(6) = 27

==>Find the rate of change/slope

f(5) = 18

f(6) = 27

Rate of change =

$\frac{f(6) - f(5)}{6-5}$

$\frac{27- 18}{6-5}$

$\frac{9}{1}$

Rate of change = 9

4 0

3 years ago

How can you prove that csc^2(θ)tan^2(θ)-1=tan^2(θ)

Oxana [17]

Answer:

Make use of the fact that as long as $\sin(\theta) \ne 0$ and $\cos(\theta) \ne 0$ :

$\displaystyle \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$ .

$\displaystyle \csc(\theta) = \frac{1}{\sin(\theta)}$ .

$\sin^{2}(\theta) + \cos^{2}(\theta) = 1$ .

Step-by-step explanation:

Assume that $\sin(\theta) \ne 0$ and $\cos(\theta) \ne 0$ .

Make use of the fact that $\tan(\theta) = (\sin(\theta)) / (\cos(\theta))$ and $\csc(\theta) = (1) / (\sin(\theta))$ to rewrite the given expression as a combination of $\sin(\theta)$ and $\cos(\theta)$ .

$\begin{aligned}& \csc^{2}(\theta) \, \tan^{2}(\theta) - 1\\ =\; & \left(\frac{1}{\sin(\theta)}\right)^{2} \, \left(\frac{\sin(\theta)}{\cos(\theta)}\right)^{2} - 1 \\ =\; & \frac{\sin^{2}(\theta)}{\sin^{2}(\theta)\, \cos^{2}(\theta)} - 1\\ =\; & \frac{1}{\cos^{2}(\theta)} - 1\end{aligned}$ .

Since $\cos(\theta) \ne 0$ :

$\displaystyle 1 = \frac{\cos^{2}(\theta)}{\cos^{2}(\theta)}$ .

Substitute this equality into the expression:

$\begin{aligned}& \csc^{2}(\theta) \, \tan^{2}(\theta) - 1\\ =\; & \cdots\\ =\; & \frac{1}{\cos^{2}(\theta)} - 1 \\ =\; & \frac{1}{\cos^{2}(\theta)} - \frac{\cos^{2}(\theta)}{\cos^{2}(\theta)} \\ =\; & \frac{1 - \cos^{2}(\theta)}{\cos^{2}(\theta)}\end{aligned}$ .

By the Pythagorean identity, $\sin^{2}(\theta) + \cos^{2}(\theta) = 1$ . Rearrange this identity to obtain:

$\sin^{2}(\theta) = 1 - \cos^{2}(\theta)$ .

Substitute this equality into the expression:

$\begin{aligned}& \csc^{2}(\theta) \, \tan^{2}(\theta) - 1\\ =\; & \cdots \\ =\; & \frac{1 - \cos^{2}(\theta)}{\cos^{2}(\theta)} \\ =\; & \frac{\sin^{2}(\theta)}{\cos^{2}(\theta)}\end{aligned}$ .

Again, make use of the fact that $\tan(\theta) = (\sin(\theta)) / (\cos(\theta))$ to obtain the desired result:

$\begin{aligned}& \csc^{2}(\theta) \, \tan^{2}(\theta) - 1\\ =\; & \cdots \\ =\; & \frac{\sin^{2}(\theta)}{\cos^{2}(\theta)}\\ =\; & \left(\frac{\sin(\theta)}{\cos(\theta)}\right)^{2} \\ =\; & \tan^{2}(\theta)\end{aligned}$ .

5 0

2 years ago