If he drives 90,000 miles a year and drives the same amount of miles each week, he drives 7,500 miles each month.
12 months are in a year. Divide 90,000 by 12
90,000/12=7,500
Answer:
Per child admission costs $8.5
Per Adult admission costs $13.5
Step-by-step explanation:
Let x be the amount of admission per child and y be the amount of admission per adult
For Tom's party:
13x+2y=137.50 (Equation 1)
For Lisa's Party:
9x+2y=105.50 (Equation 2)
Using Elimination Method:
Add Equation 1 & 2
13x+2y=137.50
9x+2y=105.50
- - = - (Change of Signs)
---------------------------------
4x+0y=34.00
x=34/4
x=8.5
Putting the value of x in Equation 2
13(8.5)+2y=137.50
110.5+2y=137.50
2y=137.50-110.50
2y=27
y=27/2
y=13.5
Answer:
a. x = 82 degrees
b. All angles less than 90 degrees are 65 degrees and 33 degrees and 82 degrees in an acute triangle.
Step-by-step explanation:
A triangle has to equal to 180 degrees, add 65 and 33 together to get 98 degrees, subtract 180 from 98 to get 82 degrees. Part B is self explanatory.
Answer:
Check the explanation.
Step-by-step explanation:
As the graph of a linear function f passes through the point (-2,-10) and has a slope of 5/2.
As the slop-intercept form is given by:

where m is the slope and b is the y-intercept.
substituting the values (-2, -10) and m = 5/2 in the slop-intercept form to determine y-intercept.






And the equation of the line in the slope-intercept form will be:

putting b = -5 and slope = m = 5/2



Determining the zero of function.
As we know that the real zero of a function is the x‐intercept(s) of the graph of the function.
so let us determine the value of x (zero of function) by setting y = 0.





Therefore, the zeros of the function will be:
if you've read those links already, you'd know what we're doing here.
we'll move the repeating part to the left-side of the dot, by multiplying by "1" and as many zeros as needed, or 10 at some power pretty much.
on 0.13 we need 100 to get 13.13.... and on 0.1234, we need 10000 to get 1234.1234....
![\bf 0.\overline{13}~\hspace{10em}x=0.\overline{13} \\\\[-0.35em] ~\dotfill\\\\ \begin{array}{|lll|ll} \cline{1-3} &&\\ 100\cdot 0.\overline{13}& = & 13.\overline{13}\\ 100\cdot x&& 13 + 0.\overline{13}\\ 100x&&13+x \\&&\\ \cline{1-3} \end{array}\implies \begin{array}{llll} 100x=13+x\implies 99x=13 \\\\ x=\cfrac{13}{99} \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ 0.\overline{1234}~\hspace{10em}x=0.\overline{1234} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%20%5Cbf%200.%5Coverline%7B13%7D~%5Chspace%7B10em%7Dx%3D0.%5Coverline%7B13%7D%0A%5C%5C%5C%5C%5B-0.35em%5D%0A~%5Cdotfill%5C%5C%5C%5C%0A%5Cbegin%7Barray%7D%7B%7Clll%7Cll%7D%0A%5Ccline%7B1-3%7D%0A%26%26%5C%5C%0A100%5Ccdot%200.%5Coverline%7B13%7D%26%20%3D%20%26%2013.%5Coverline%7B13%7D%5C%5C%0A100%5Ccdot%20x%26%26%2013%20%2B%200.%5Coverline%7B13%7D%5C%5C%0A100x%26%2613%2Bx%0A%5C%5C%26%26%5C%5C%0A%5Ccline%7B1-3%7D%0A%5Cend%7Barray%7D%5Cimplies%20%5Cbegin%7Barray%7D%7Bllll%7D%0A100x%3D13%2Bx%5Cimplies%2099x%3D13%0A%5C%5C%5C%5C%0Ax%3D%5Ccfrac%7B13%7D%7B99%7D%0A%5Cend%7Barray%7D%0A%5C%5C%5C%5C%5B-0.35em%5D%0A%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%0A0.%5Coverline%7B1234%7D~%5Chspace%7B10em%7Dx%3D0.%5Coverline%7B1234%7D%0A%5C%5C%5C%5C%5B-0.35em%5D%0A~%5Cdotfill%20)
