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4 years ago

The figure below shows a line graph and two shaded triangles that are similar:

Mathematics

2 answers:

Vlad1618 [11]4 years ago

7 0

The answer is the first one.

The line only has one slope, which in this case is -1/3

maks197457 [2]4 years ago

4 0

The answer is -1/3, could you please give me brainliest im trying to rank up.

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Given the Cos(x) = -2/3, then the Sec(x) = ? What is the answer

ivanzaharov [21]

Answer:

$sec x = - \frac{3}{2}$

Step-by-step explanation:

$cos x = \frac{-2}{3}\\\\sec x = \frac{1}{cosx}\\\\Therefore, sec x = \frac{3}{-2} = \frac{-3}{2}$

6 0

3 years ago

Read 2 more answers

Standard for ola (y + 3)2 = -12(x - 2), iden 1. D n111

Vedmedyk [2.9K]

Answer:

See below

Step-by-step explanation:

Your question is a bit unclear, so I'm going to assume you want the value of y or x:

(y+3)^2=-12(x-2)

-(y+3)^2/12=x-2

-(y+3)^2/12+2=x

(y+3)^2=-12(x-2)

y+3=sqrt(-12x+24)

y=sqrt(-12x+24)-3

5 0

3 years ago

Choose the product of -3y?(-4y + y-9).

Trava [24]

Answer:

Step-by-step explanation:

D. 12y5 – 3y + 27y2

8 0

3 years ago

Part 1: What mistake did AJ make in the graph?

grigory [225]

Answer:

Part 1) AJ drawn the parabola opening upwards, instead of drawing it opening downwards

Part 2) see the explanation

Step-by-step explanation:

Part 1) What mistake did AJ make in the graph?

we have

$f(x)=-(x+2)^2-1$

This is the equation of a vertical parabola written in vertex form

The parabola open downward (because the leading coefficient is negative)

The vertex represent a maximum

The vertex is the point (-2,-1)

therefore

AJ drawn the parabola opening upwards, instead of drawing it opening downwards

Part 2) Evaluate any two x-values (between -5 and 5) into AJ's function. Show your work. How does your work prove that AJ made a mistake in the graph?

take the values x=-4 and x=4

For x=-4

substitute the value of x in the quadratic equation

$f(x)=-(-4+2)^2-1\\f(x)=-5$

For x=4

substitute the value of x in the quadratic equation

$f(x)=-(4+2)^2-1\\f(x)=-37$

According to AJ's graph for the value of x=-4 the function should be positive, however it is negative and for the value of x=4 the function should be positive and the function is negative

therefore

AJ made a mistake in the graph

8 0

3 years ago

A triangle has a perimeter of 28 cm. If the three sides of the triangle are n, 3n − 4, and 5n − 13, what is the length of each s

jeka57 [31]

28=9n-17 45=9n 5=n One side is 5 One side is 11 One side is 12

8 0

4 years ago