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3 years ago

Find the range of f(x)=3g-5 for domain (-1.5,2,4)

1 answer:

8 0

I guess f(x) should be f(g)

f(g) = 3g-5

f(-1.5) = 3(-1.5) -5 =-9

f(2) = 3(2) - 5 = 1

f(4) = 3(4) -5 7

In short you replace the value of g (the domaine) to find f(x) the range

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Divide 250 in the ratio of 1 : 4

Aleksandr [31]

1 + 4 = 5 (5 parts)
£250/5 = 50 
1 x £50 = 50 
4 x £50 = 200 
so the answer is 50:200

6 0

3 years ago

Read 2 more answers

The endpoints of a line segment are at the coordinates (–6, 3, 4) and (4, –1, 2). What is the midpoint of the segment?

natulia [17]

Answer:

$( - 1,1,3 )$

Step-by-step explanation:

Use the midpoint formula:

$(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}, \frac{z_1+z_2}{2} )$

We want to find the midpoint of the line

that has endpoints.

$(\frac{ - 6+4}{2},\frac{3+ - 1}{2}, \frac{4+2}{2} )$

We simplify to get;

$(\frac{ - 2}{2},\frac{2}{2}, \frac{6}{2} )$

This finally gives:

$( - 1,1,3 )$

The correct answer is A.

7 0

3 years ago

Could someone help me with this question? i don't know how to really work this out Order Of Operations

lozanna [386]

Answer:

Step-by-step explanation:

2[45/(11-8)^2]

=2[45/(3)^2]

=2[45/9]

=2[5]

=10

3 0

3 years ago

Help! I have looked at so many lessons on writing exponential functions from a graph and nothing has helped.

Over [174]

Answer:

y = $8(-\frac{1}{2})^x$

Step-by-step explanation:

Let the equation of the exponential function is,

y = a(b)ˣ

Since the graph of this function passes through two points $(4, \frac{1}{2})$ and (3, -1)

For the point $(4, \frac{1}{2})$ ,

$\frac{1}{2}=a(b)^4$ ---------(1)

For second point (3, -1),

-1 = a(b)³ ---------(2)

Divide equation (2) by equation (1),

$\frac{\frac{1}{2}}{-1}$ = $\frac{a(b)^4}{a(b)^3}$

$-\frac{1}{2}=b^{(4-3)}$

$-\frac{1}{2}=b$

From equation (2)

-1 = $a(-\frac{1}{2})^3$

-1 = $-\frac{1}{8}a$

a = 8

Therefore, equation of the exponential function will be,

y = $8(-\frac{1}{2})^x$

4 0

3 years ago

Combine like terms to create an equivalent expression.

Lelu [443]

Answer:

$\frac{1}{6}$ p - $\frac{4}{5}$

Step-by-step explanation:

collect like terms

- $\frac{2}{3}$ p + $\frac{5}{6}$ p + $\frac{1}{5}$ - 1

= - $\frac{2(2)}{2(3)}$ p + $\frac{5}{6}$ p + $\frac{1}{5}$ - $\frac{5}{5}$

= - $\frac{4}{6}$ p + $\frac{5}{6}$ p - $\frac{4}{5}$

= $\frac{1}{6}$ p - $\frac{4}{5}$

5 0

2 years ago