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4 years ago

For the given domain value fill in the Y value for the equation Y=2/3x-3

Mathematics

1 answer:

Sveta_85 [38]4 years ago

4 0

Y=2/3 times x/1-3
=y=2x/3-3
y=2/3x-3

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5 and 1/3 × 3 and 3/4

tresset_1 [31]

We'll first off convert the mixed fractions to "improper", and then multiply.

$\bf \stackrel{mixed}{5\frac{1}{3}}\implies \cfrac{5\cdot 3+1}{3}\implies \stackrel{improper}{\cfrac{16}{3}}
\\\\\\
\stackrel{mixed}{3\frac{3}{4}}\implies \cfrac{3\cdot 4+3}{4}\implies \stackrel{improper}{\cfrac{15}{4}}
\\\\\\
\cfrac{16}{3}\times \cfrac{15}{4}\implies \cfrac{15}{3}\times \cfrac{16}{4}\implies 5\times 4\implies 20$

6 0

3 years ago

For the functions f(x) = 2x – 3 and g(x) = 5x + 5, find (f • g)(x).

Nimfa-mama [501]

Answer:

f(g(x)) = 10x + 7

Step-by-step explanation:

(f . g)(x) = f(g(x)) is the composition of the functions f(x) and g(x).

f(g(x)) means that g(x) becomes the input for f.

Now, f(x) = 2x - 3 and g(x) = 5x + 5.

f(g(x)) = f(5x + 5)

= 2((5x + 5) - 3)

= 10x + 10 - 3

= 10x + 7

f(g(x)) = 10x + 7 is the required answer.

5 0

4 years ago

(WORTH 10 POINTS PLEASE ANSWER ASAP PLEASEEEEEEE)

Brilliant_brown [7]

Answer:

2(10+6)=32

Step-by-step explanation:

8 0

4 years ago

Read 2 more answers

A satellite in a circular orbit 1250 kilometers above Earth makes one complete revolution every 110 minutes. What is its linear

Ivenika [448]

Answer:

$\large \boxed{\text{approximately 435.7 km/min}}$

Step-by-step explanation:

1. Angular speed

The angular speed ω is the angle θ swept out by the satellite in a given time t.

$\omega = \dfrac{\theta}{t} = \dfrac{2\pi}{\text{110 min}}$

2. Linear speed

The formula for the linear speed v is

v = rω, where

r = the distance from the centre of the Earth = 6378 km + 1250 km = 7628 km

$\begin{array}{rcl}v & = & r\omega\\& = & \text{7268 km} \times \dfrac{2\pi}{\text{110 min}}\\\\& \approx & \textbf{435.7 km/min}\\\end{array}\\\text{The linear speed of the satellite is $\large \boxed{\textbf{approximately 435.7 km/min}}$}$

7 0

3 years ago

A student’s construction of the perpendicular bisector of is shown below. What is the student’s error?

Elena-2011 [213]

The student did not make the opening of each arc drawn from S and T greater than half the length of the line ST. When drawing a perpendicular bisector, the arcs should go past the center of the line segment.

4 0

3 years ago