All categories

3 years ago

PLEASE HELP ITS DUE TMR N I JUST NEED A ANSWER

Mathematics

1 answer:

Daniel [21]3 years ago

7 0

The second answer because addition is commutative.

You might be interested in

Question 2 (1 point)

konstantin123 [22]

He does 60 push ups

4 0

3 years ago

Read 2 more answers

Write an equation of the line in point-slope form that passes through the given points (-3,2) (5,5)

daser333 [38]

I think it is: y=0.375x+3.125

a=(5-2)/(5- -3)=3/8=0.375

Find b

(5)=0.375(5)+b
(5)=1.875+b
5-1.875=b
b=3.125

3 0

3 years ago

PLEASE NO LINKS I LOST SO MANY POINTS

Ann [662]

Answer:

We will split the trapezoid into right angle triangle and rectangle

Traingle with base 3cm and height 10 - 6 = 4cm

Area of traingle

= $\frac{1}{2}*base *height = \frac{1}{2} * 3 * 4 = 6cm^{2}$

Area of rectangle with sides 6cm and 3cm

= $6 * 3 = 18cm^{2}$

Therefore, area of the trapezoid

= $6+18 = 24cm^{2}$

Area of the semi circle with radius 3/2 = 1.5cm

= $\frac{1}{2}\pi r^{2} = \frac{1*\pi *1.5*1.5}{2} = 1.125\pi cm^{2}$

$Total area = 24 + 1.125\pi =27.53cm^{2}$

8 0

3 years ago

For a certain model of car the distance d required to stop the vehicle if it is traveling at v mi/h is given by the formula d =

enyata [817]

Given that <span>For a certain model of car the distance $d$ required to stop the vehicle if it is traveling at $v$ mi/h is given by the formula $d=v+\frac{v^2}{20}, where [tex]d$ is measured in feet.

If Kerry wants her stopping distance not to exceed 75 ft, then the range of speeds (in mi/h) can she travel is obtained as follows:

$v+ \frac{v^2}{20} \leq75 \\ \\ 20v+v^2\leq1,500 \\ \\ v^2+20v+100\leq1,500+100 \\ \\ (v+10)^2\leq1,600 \\ \\ |v+10|\leq\sqrt{1,600} \\ \\ -40\leq v+10\leq40 \\ \\ -40-10\leqv\leq v\leq40-10 \\ \\ -50\leq v\leq30$

Therefore, the range of speed she can travel is $v\leq30\ mi/hr$
</span>

8 0

3 years ago

What is the slope of the line represented by the equation y = y equals 4 Over 5

OLEGan [10]

$\bf y= \stackrel{\stackrel{m}{\downarrow }}{\cfrac{4}{5}}x-3\qquad \impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}$

4 0

3 years ago

Read 2 more answers