All categories

3 years ago

Please help!!! If h(x)= -3x-10, find h(-3) Please give an explanation too. Thank you.

Mathematics

2 answers:

tankabanditka [31]3 years ago

8 0

Answer:

-1

Step-by-step explanation:

If h(x)= -3x-10 and we want to solve for h(-3) then we have to plug in -3 for all x variables. h(-3)= -3(-3)-10, then solve using PEMDAS.

First multiple h(-3)= 9-10, multiple -3*-3 to get 9.

Then subtract, h(-3)=-1

dalvyx [7]3 years ago

5 0

Answer:

h(-3)=-1

Step-by-step explanation:

Replace all x in h(x)= -3x-10 by -3:

h(-3)=-3(-3)-10

Now multiply -3 and -3:

h(-3)=9-10

Simplify 9 and -10 (aka add 9 and -10 together):

h(-3)=-1

Answer: h(-3)=-1

Hope this helps!

You might be interested in

What is the minimum number of points through which a line is drawn

Rudik [331]

The <span>minimum number of points through which a line is drawn is 2 points please let me know if im wrong</span>

8 0

4 years ago

The Tennessee Titans scored 22 points in week 5 and 35 points in week 6. Find the percent of increase.

Shalnov [3]

Answer:

59.09%

Step-by-step explanation:

35-22=13

13/2= .59090909=59.09%

3 0

3 years ago

Compare the exponential model of population growth with the logistic model.

Pavlova-9 [17]

No picture of graph no answer sorry

8 0

3 years ago

The position of an object moving along an x axis is given by x = 3.24 t - 4.20 t2 + 1.07 t3, where x is in meters and t in secon

Pavel [41]

Answer and explanation:

Given : The position of an object moving along an x axis is given by $x=3.24t-4.20t^2+1.07t^3$ where x is in meters and t in seconds.

To find : The position of the object at the following values of t :

a) At t= 1 s

$x(t)=3.24t-4.20t^2+1.07t^3$

$x(1)=3.24(1)-4.20(1)^2+1.07(1)^3$

$x(1)=3.24-4.20+1.07$

$x(1)=0.11$

b) At t= 2 s

$x(t)=3.24t-4.20t^2+1.07t^3$

$x(2)=3.24(2)-4.20(2)^2+1.07(2)^3$

$x(2)=6.48-16.8+8.56$

$x(2)=-1.76$

c) At t= 3 s

$x(t)=3.24t-4.20t^2+1.07t^3$

$x(3)=3.24(3)-4.20(3)^2+1.07(3)^3$

$x(3)=9.72-37.8+28.89$

$x(3)=0.81$

d) At t= 4 s

$x(t)=3.24t-4.20t^2+1.07t^3$

$x(4)=3.24(4)-4.20(4)^2+1.07(4)^3$

$x(4)=12.96-67.2+68.48$

$x(4)=14.24$

(e) What is the object's displacement between t = 0 and t = 4 s?

At t=0, x(0)=0

At t=4, x(4)=14.24

The displacement is given by,

$\triangle x=x(4)-x(0)$

$\triangle x=14.24-0$

$\triangle x=14.24$

(f) What is its average velocity from t = 2 s to t = 4 s?

At t=2, x(2)=-1.76

At t=4, x(4)=14.24

The average velocity is given by,

$\triangle x=x(4)-x(2)$

$\triangle x=14.24-(-1.76)$

$\triangle x=14.24+1.76$

$\triangle x=16$

4 0

3 years ago

I need help answering this question.

Darina [25.2K]

Her credits need to be AT LEAST EQUAL TO OR GREATER THAN 144 (credit hours) so we can eliminate choices B and D.

She already completed 4 semesters in which she receives 15 credit per semester.

This expression can be written as 4(15).

She needs to do a certain remaining hours of credit which can be represented by c and only c without any other coefficient.

So, the most reasonable choice here is A.

7 0

3 years ago

Read 2 more answers