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3 years ago

Estimate the solution to the following system of equations by graphing.

Mathematics

1 answer:

stealth61 [152]3 years ago

8 0

For graphing the following equations I have my answers as

(-0.5,0.75)

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Which two values of x are roots of the polynomial below?

Julli [10]

Answer:

B. and C.

Step-by-step explanation:

1. Set the polynomial equal to zero:

x² + 5x + 7 = 0

2. Plug the given values of a, b, and c into the quadratic formula:

$x=\frac{-5+-\sqrt{5^{2}-4(1)(7) } }{2(1)}$

3. Solve the square root:

$x=\frac{-5+-\sqrt{25-4(1)(7)} }{2(1)}$

$x=\frac{-5+-\sqrt{25-28} }{2(1)}$

$x=\frac{-5+-\sqrt{-3} }{2(1)}$

4. Simplify the rest:

$x=\frac{-5+-\sqrt{-3} }{2}$

5. Separate the solutions:

$x=\frac{-5+\sqrt{-3} }{2} , x=\frac{-5-\sqrt{-3} }{2}$

Thus, options B and C are the answers.

hope this helps!

4 0

2 years ago

Approximate the square root to the nearest whole number 142

slamgirl [31]

The square root of 142 is 11.916

7 0

3 years ago

shira was riding the screamer when it broke down. her seat was 53 jorizontal feet from the central support pole. What was her se

ki77a [65]

Answer:

The possible solutions are 58 degree, 122 degrees, 302 degrees and 238 degrees

Step-by-step explanation:

Please see the attachment

From the figure, it is evident that

cos AOB = OB/OA = 53/100 = 0.53

Thus, cos-1 (0.53) = 58 degrees

AOB = 58 degree

Also, there are can be other values depending on the quadrants

COB = 180 -58 = 122 degrees

FOB = 360 -58 = 302 degrees

EOB = 180 + 58 = 238 degrees

5 0

3 years ago

What fraction of 60 is… Chapter Reference 24

sergij07 [2.7K]

Answer: 3/5

Explanation: 60/100 = 6/10 = 3/5

7 0

3 years ago

A polynomial function has a zero at 4 (multiplicity 3) and 0 (multiplicity 1). Write a function in standard form that can repres

VLD [36.1K]

Answer:

(x^4) - 64x

Step-by-step explanation:

zeros at 4 and 0 are the roots so can be factor as x*(x-4) and the multiplicity tels to what power you have to raise those factors

$x^{1} *(x-4)^{3}$ becuase multiplicity of 1 for root at 0, and multiplicity of 3 at root 4

now to write this in standard form

$x*(x-4)^{3} = \\x*(x-4)(x^{2} +4x+16)= \\(x^{2} -4x)(x^{2} +4x+16) = \\x^{4} +4x^{3} +16x^{2} -4x^{3} -16x^{2} -64x = \\x^{4} -64x$

8 0

2 years ago