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Marta_Voda [28]
3 years ago
12

Which is the approximate solution for the system of equations x + 5= 10 and 3x + y = 1?

Mathematics
2 answers:
Semmy [17]3 years ago
6 0

Answer:

Step-by-step explanation:

First solve the first equation for x:  x = 5.  Next, substitute this 5 for x in the second equation:  3(5) + y = 1, or 15 + y = 1, or y = -14.  The solution is (5, -14).

zheka24 [161]3 years ago
3 0

Answer: The approximate solution is (-0.36, 2.08).

Given system of equations : x+5y=10 and 3x+y=1

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fred is playing tennis fo every 2 of his serves that land in 4 serves land out if he hit 26 serves in how many serves landed out
ki77a [65]

Answer:

52 serves landed out

Explanation:

If you divide the number of serves in by 2 ( number that he lands in for every 6 serves) and then multiply by 4 ( number that land out for every 6 serves), you get 52 serves out.

3 0
4 years ago
Please answer asap!!!!!
kherson [118]

Step-by-step explanation:

The equation of a circle can be the expanded form of

\large \text{$(x-a)^2+(y-b)^2=r^2$}(x−a)

2

+(y−b)

2

=r

2

where rr is the radius of the circle, (a,\ b)(a, b) is the center of the circle, and (x,\ y)(x, y) is a point on the circle.

Here, the equation of the circle is,

\begin{gathered}\begin{aligned}&x^2+y^2+10x-4y-20&=&\ \ 0\\ \\ \Longrightarrow\ \ &x^2+y^2+10x-4y+25+4-49&=&\ \ 0\\ \\ \Longrightarrow\ \ &x^2+y^2+10x-4y+25+4&=&\ \ 49\\ \\ \Longrightarrow\ \ &x^2+10x+25+y^2-4y+4&=&\ \ 49\\ \\ \Longrightarrow\ \ &(x+5)^2+(y-2)^2&=&\ \ 7^2\end{aligned}\end{gathered}

⟹

⟹

⟹

⟹

x

2

+y

2

+10x−4y−20

x

2

+y

2

+10x−4y+25+4−49

x

2

+y

2

+10x−4y+25+4

x

2

+10x+25+y

2

−4y+4

(x+5)

2

+(y−2)

2

=

=

=

=

=

0

0

49

49

7

2

From this, we get two things:

\begin{gathered}\begin{aligned}1.&\ \ \textsf{Center of the circle is $(-5,\ 2)$.}\\ \\ 2.&\ \ \textsf{Radius of the circle is $\bold{7}$ units. }\end{aligned}\end{gathered}

1.

2.

Center of the circle is (−5, 2).

Radius of the circle is 7 units.

Hence the radius is 7 units.

4 0
3 years ago
Una avenida está siendo asfaltada por etapas. En la primera etapa se asfaltó la mitad; en la segunda, la quinta parte, y en la t
s344n2d4d5 [400]

Answer:

Sabemos que:

L es el largo de la avenida.

En la primer etapa se asfalto la mitad, L/2, entonces lo que queda por asfaltar es:

L - L/2 = L/2.

En la segunda etapa se asfalto la quinta parte, L/5, entonces lo que queda por asfaltar es:

L/2 - L/5 = 5*L/10 - 2*L/10 = (3/10)*L

En la tercer etapa se asfalto la cuarta parte del total, L/4, entonces lo que queda por asfaltar es:

(3/10)*L - L/4 = 12*L/40 - 10L/40 = (2/40)*L

Y sabemos que este ultimo pedazo que queda por asfaltar es de 200m:

(2/40)*L = 200m

L = 200m*(40/2) = 4,000m

7 0
3 years ago
What is the answers
postnew [5]
2,4,6,8,10,0,0,0,0,0,1,4,9,16,25,1,2,3,4,5
4 0
3 years ago
Read 2 more answers
Angles T and V are country that angle T has a measure of (2X +10) angle B has a measure of 48° what is the value of X
enyata [817]
If T and V are complementary angles, their sum is 90°.

   V + T = 90°
   48° + (2X+10)° = 90° . . . . . . . substitute given information
   2X + 58 = 90 . . . . . . . . . . . . .. collect terms
   2X = 32 . . . . . . . . . . . . . . . . .. subtract 58
   X = 16 . . . . . . . . . . . . . . . . . .. divide by 2

The value of X is 16.
4 0
3 years ago
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