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3 years ago

Select all the points that are on the graph of the function defined by y = 2x + 3

Mathematics

2 answers:

ale4655 [162]3 years ago

4 0

Answer:

b, d, e, g are the points

Kaylis [27]3 years ago

3 0

All you need to do to solve this is plug in x and y values.

a) -8 + 3 = -5, not 3
b) -2 + 3 = 1
c) 0 + 3 = 3, not 0
d) 4 + 3 = 7
e) 6 + 3 = 9
f) 6 + 3 = 9, not 12
g) 9 + 3 = 12

The correct answers are B, D, E, G

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Please Help!

MAVERICK [17]

For a translation of 4 units right, the graph will become -3(x-4). the 4 is negative to indicate that the translation is right.
For a vertical stretch of 4, the graph will become 4x-3(x-4)=-12(x-4).

With this information, g(x)=-12(x-4)

5 0

3 years ago

Find all real and imaginary solutions to the equation x^3-4x^2+x-4=0

lawyer [7]

Step-by-step explanation:

STEP 1

Equation at the end of step 1

(((x3) - 22x2) + x) - 4 = 0

STEP 2

Checking for a perfect cube

2.1 x3-4x2+x-4 is not a perfect cube

5 0

3 years ago

The steps to convert fractions into decimals

Allisa [31]

Answer:

Just simplify!!

EXample: 11 over 8

1 and 3 over 8

=1.375

7 0

3 years ago

What are the points

Goryan [66]

Answer:

There are 9 dots. So 9.

Step-by-step explanation:

Just take my word for it.

7 0

3 years ago

Match each of the trigonometric expressions below with the equivalent non-trigonometric function from the following list. Enter

Levart [38]

Answer:

Match each of the trigonometric expressions below with theequivalent non-trigonometric function from the following list.Enter the appropiate letter(A,B, C, D or E)in each blank

A . tan(arcsin(x/8))

B . cos (arsin (x/8))

C. (1/2)sin (2arcsin (x/8))

D . sin ( arctan (x/8))

E. cos (arctan (x/8))

These are the spaces to fill out :

.. ..........x/64 (sqrt(64-x^2))

.............x/sqrt(64+x^2)

.............sqrt(64-x^2)/8

..............x/sqrt(64-x^2)

..............8/sqrt(64+x^2)

A. ........tan(arcsin(x/8)) =......x/sqrt(64-x^2)

B . cos (arsin (x/8)) ....sqrt(64-x^2)/8

Step-by-step explanation:

To solve this we have to find the missing sides to each of the triange discribed in prenthesis thus

A we have the sides of the triangle given by x, 8 and $\sqrt{8^{2} - x^{2} }$ or $\sqrt{64 - x^{2} }$

thus tan(arcsin(x/8)) = $\frac{x}{\sqrt{64 - x^{2} }}$ =

Therefore ........tan(arcsin(x/8)) =......x/sqrt(64-x^2)

Here we have cos = adjacent/hypotenuse where adjacent side is $\sqrt{64 - x^{2} }$ and hypothenuse = 8 we have $\sqrt{64 - x^{2} }$ /8

B . cos (arsin (x/8)) ....sqrt(64-x^2)/8

4 0

2 years ago