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Rashid [163]
3 years ago
7

Please help me Find x, y, and z

Mathematics
1 answer:
Evgesh-ka [11]3 years ago
5 0

Answer:

X 90 z 90 y 90

Step-by-step explanation:

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Please help with algebra class I’m struggling
adell [148]

Answer:

Step-by-step explanation:

For what mileages will company A (which has a flat rate of $111) be less than the cost for company B

Company A < Company B

              111 < 75 + 0.80m

solving

       111 - 75 < 75 - 75 + 0.80m

              36 < 0.80m

     36/0.80 < 0.80m/0.80

as we are dividing by a positive number, we don't have to change the direction of the < sign.

              45 < m

      or     m > 45

so if Ali intends to drive more than 45 miles, Company A's vehicle will be less expensive to rent

5 0
3 years ago
Find the distance between (2,3) and (5,8)
Orlov [11]
Use pythagorean theorem
a^2+b^2=c^2

first find the difference of between the 2 points to find the legnth of the legs

distance between 2 and 5=3
distance between 3 and 8=5 so


3^2+5^2=c^2
9+25=c^2
34=c^2
√34=c

the ligit formula is
c= \sqrt{(x1-x2)^2+(y1-y2)^2}

6 0
3 years ago
(4,-2) and (-2,-2) write the equation.​
bixtya [17]

the equation is Y=-2/3x+2/3

8 0
4 years ago
Train A arrives at central station on the hour and every 12 minutes. Train B arrives on the hour and every 15 minutes. When do b
Lapatulllka [165]
<span>2________24___________30
3________36___________45
4________48___________60
5________60, found common minute count. 

</span><span>60 minutes.</span><span>__________A____________B
__________0____________0
1________12___________15
</span>
3 0
3 years ago
Suppose a punter kicks a football so that the upward component of its velocity is 80 feet per second. If the ball is 3 feet off
Alexxandr [17]
So hmm check the picture below

\bf \qquad \textit{initial velocity}\\\\ h = -16t^2+v_ot+h_o \qquad \text{in feet}\\&#10;\\&#10;v_o=\textit{initial velocity of the object}\\&#10;h_o=\textit{initial height of the object}\\&#10;h=\textit{height of the object at "t" seconds}\\\\&#10;-----------------------------\\\\&#10;h(t)=3+80t-16t^2\iff h(t)=-16t^2+80t+3

a)

well, clearly is 80 ft/s

b)

when t = 1? well 80(1)

c)

in the picture, x-axis is the time and y-axis is the height
so, it reaches its maximum at the vertex, after "x" seconds

\bf \begin{array}{lcccll}&#10;h(t)=&-16t^2&+80t&+3\\&#10;&\uparrow &\uparrow &\uparrow \\&#10;&a&b&c&#10;\end{array}\qquad &#10;\left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad  {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)

so it reached the vertex after \bf -\cfrac{{{ b}}}{2{{ a}}}\quad seconds

d)

the maximum height of the ball is \bf {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\quad feet

8 0
3 years ago
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