Can you help me with this

Kinda confused on this one so it would be helpful to answer this for 30 points! And marked!

seropon [69]

Answer:

acute

Step-by-step explanation:

straight is just a straight line, 180 degrees perhaps

right is exactly 90 degrees

obtuse is over 90 degrees

5 0

3 years ago

It cost Makayla $2.80 to send 28 text messages. How much would it cost to send 163 text messages?

RUDIKE [14]

Answer:

$16.30

Step-by-step explanation:

Hope this helps! Have a great day!!

8 0

2 years ago

Compare the two functions to see which has a greater rate of change. Function 1

AleksAgata [21]

Function 2 has greater rate of change

Step-by-step explanation:

In order to find the rate of change we have to covert the given linear functions in slope-intercept form

Function 1:

$-2x+5y = 10$

Adding 2x on both sides

$-2x+5y+2x = 2x+10\\5y = 2x+10$

Dividing both sides by 5

$\frac{5y}{5} = \frac{2x+10}{5}\\y = \frac{2}{5}x +\frac{10}{5}\\y = \frac{2}{5}x+2$

Let m1 be rate of change of function 1:

$m_1 = \frac{2}{5}$

Function 2:

$-6x+3y = 18$

Adding 6x on both sides

$-6x+3y+6x = 6x+18\\3y = 6x+18$

Dividing both sides by 3

$\frac{3y}{3} = \frac{6x+18}{3}\\y = \frac{6x}{3} + \frac{18}{3}\\y= 2x+6$

Let m2 be the slope of function 2

$m_2 = 2$

As we can see that

$m_2>m_1$

Function 2 has greater rate of change

Keywords: Linear functions, slope

Learn more about functions at:

#LearnwithBrainly

7 0

3 years ago

H=-16t^2 + 104t + 56

WITCHER [35]

Answer:

The graph in the attached figure

Step-by-step explanation:

we have

$h=-16t^{2}+104t+56$

Using a graphing tool

see the attached figure

This is the equation of a vertical parabola open downward

The vertex is the point (3.25,225) (a maximum)

The zero's of the function or the t-intercepts are the points (-0.5,0) and (7,0)

The h-intercept of the function is the point (0,56)

The domain of the function are all real numbers

The range of the function $h\leq225$ ----> All real numbers less than or equal to 225

8 0

4 years ago

A ball is thrown from an initial height of 2 feet with an initial upward velocity of 33/fts. The ball's height h (in feet) after

RoseWind [281]

Answer:

t=1.283 seconds and

0.779 seconds

Step by step Explanation:

Given: h=18 ft

The given equation is h=2+33t-16t²

Then if we substitute the value of given h, h=18 ft into the given equation we have,

18=2+33t-16t²

Then if we re- arrange we have

16t²−33t+16=0

We can see that the above quadratic equation is in standard form, with a=16, b=33 and c=16 then we can use quadratic formula in solving it which is

t= −(−33±√[(−33) ²−4×16×16)]/(2×16)

= [33±√[1089−1024]/(32)

= [33±√[65]/(32)

=1.283 or 0.779 seconds

the two real roots , of the quadratic are:

1.283 and

0.779 seconds

t= 1.283 or 0.779 seconds

Hence, the ball is at 18 feet with height 0.779seconds after it has been thrown up and,

and is at 21 feet with height 1.283 seconds after after thrown down

4 0

3 years ago