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Tresset [83]
3 years ago
7

How many solutions exist for the following system of equations: 4x + y = 7 and 3x - y = 0 a. no solution b. one solution c. infi

nitely many d. cannot be determined
Mathematics
1 answer:
Zanzabum3 years ago
6 0
4x+y=7 \\
\underline{3x-y=0} \\
4x+3x=7+0 \\
7x=7 \ \ \ |\div 7 \\
x=1 \\ \\
3x-y=0 \\
3 \times 1-y=0 \\
3-y=0 \ \ \ |+y \\
y=3 \\ \\
(x,y)=(1,3)

The answer is B. one solution.
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Please help me out I need help so bad
wel

the answer is definitly C

5 0
3 years ago
What is the distance between (-6, 8) and (-3,9)?
vivado [14]

Answer:

is \sqrt{10}

Step-by-step explanation:

var x = +3 (one side of right angled triangle)

var y = +1 (the other side of the right angled triangle)

Use the Pythagoras' Theorem.

3² + 1² = 10

6 0
3 years ago
10 3/9- 10 2/5 Im to lazy to do it please help me out.
coldgirl [10]

Answer:

\boxed{ \ -  \frac{3}{45} }

Step-by-step explanation:

10 3/9 - 10 2/5

we need to convert mixed fractions to improper fractions

10 \frac{3}{9}  =  \frac{10 \times 9 + 3}{9}  =  \frac{93}{9}

10 \frac{2}{5}  =  \frac{10 \times 5 + 2}{5}  =  \frac{52}{5}

Now, we have 93/9 - 52/5

We need to equate the denominators, before subtracting them.

\frac{93}{9}  -  \frac{52}{5}  =  \frac{93 \times 5}{45}  -  \frac{52 \times 9}{45}

\frac{465}{45}  -  \frac{468}{45}

\frac{ - 3}{45}

\boxed{ -  \frac{3}{45} }

The result of 10 3/9- 10 2/5 is - 3/45

8 0
2 years ago
752 second graders went to
Sophie [7]
752 second graders went to
see Hansel and Gretel on
Friday. 25% of the students
went to the bathroom during
the movie. How many kids
went to the bathroom?
8 0
3 years ago
Read 2 more answers
What are the real and complex solutions of the polynomial equation? x^3-8=0. with imaginary numbers
seraphim [82]

Answer:

Solutions are 2,  -1 +  0.5 sqrt10 i  and -1 - 0.5 sqrt10 i

or 2,  -1 +  1.58 i  and -1 - 1.58i

(where the last 2 are equal to nearest hundredth).


Step-by-step explanation:

The real solution is x = 2:-

x^3 - 8 = 0

x^3 = 8

x = cube root of 8 = 2

Note that a cubic equation must have  a total of 3 roots ( real and complex in this case).  We can find the 2 complex roots by using the following identity:-

a^3 - b^3 = (a - b)(a^2 + ab + b^2).

Here  a = x and b = 2 so we have

(x - 2)(x^2 + 2x + 4) = 0

To find the complex roots we solve x^2 + 2x + 4 = 0:-

Using the quadratic formula x = [-2 +/- sqrt(2^2 - 4*1*4)] / 2

= -1 +/- (sqrt( -10)) / 2

= -1 +  0.5 sqrt10 i  and -1 - 0.5 sqrt10 i

4 0
2 years ago
Read 2 more answers
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