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vitfil [10]
4 years ago
13

Ociel has designed a square mural that measures 10 feet on each side. Bill has also designed a square mural, but his measures y

feet shorter on each side. How much smaller than Ociel's mural is Bill's mural? Explain
Mathematics
1 answer:
WINSTONCH [101]4 years ago
6 0
Ocel’s 10x10=100
Bill’s (10 - y) (10 - y)

I don’t know which answer your teacher wants
100 - [(10-y)(10-y)]
100 - (100 -20y + y^2}
20y - y^2
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Find the solution of this system of equations <br> -6y=-50-4x <br> 5x-6y=-49
Maslowich

Answer:

<h2>(1,\ 9)</h2>

Step-by-step explanation:

Rewrite the first equation

-6y=-50-4x\\\\4x-6y=-50

Now we have the following system of equations

4x-6y=-50\\5x-6y=-49

To solve the system of equations multiply the first equation by -1 and add it to the second equation

-1*4x-(-1)*6y=-50*(-1)\\\\-4x+ 6y=50

-4x+ 6y=50

        +

5x-6y=-49

-----------------------------------

x + 0 =1\\\\x=1

Now substitute the value of x in any of the two equations and solve for y

5(1)-6y=-49

5-6y=-49

-6y=-49-5

-6y=-54

y=\frac{54}{6}

y=9

The solution is:

(1,\ 9)

6 0
3 years ago
Read 2 more answers
Find the measure for angle f<br><br> 46<br> 34<br> 122<br> 58
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Answer:

\angle f = 122\degree

Step-by-step explanation:

\angle d = 58\degree (vertical angles)

\angle f+\angle d= 180\degree

(Interior angles on the same side of transversal)

\angle f+ 58\degree = 180\degree

\angle f = 180\degree-58\degree

\huge \orange {\boxed {\angle f = 122\degree}}

5 0
3 years ago
There is a circle inscribed in an equilateral triangle, and there is another equilateral triangle inscribed in the circle. The r
Ksenya-84 [330]

Answer:

Assuming that the circle bissects the triangles equal sides at their midpoints, the ratio of the larger to smaller triangle should be close to 1/3

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Write the equation 3x2 + 2y2 + 42x - 8y + 149 = 0 in the standard form
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Its an ellipse. 3 (x+7)^2+2 (y-2)^2-6 = 0
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Using an ordered alphabet of 26 letters, how many ways are there to choose a set of six letters such that no two letters in the
mestny [16]
To find our solution, we can start off by creating a string of 27 boxes, all followed by the letters of the alphabet. Underneath the boxes, we can place 6 pairs of boxes and 15 empty boxes.The stars represent the six letters we pick. The empty boxes to the left of the stars provide the "padding" needed to ensure that no two adjacent letters are chosen. We can create this - 
\binom {21} {6}= \frac{21*20*19*18*17*16}{6*5*4*3*2*1} =21*19*17*8=54264
Thus, the answer is that there are \boxed{54264} ways to choose a set of six letters such that no two letters in the set are adjacent in the alphabet. Hope this helped and have a phenomenal New Year! <em>2018</em>
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