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3 years ago

What is the largest positive integer value of m such that the equation

1 answer:

monitta3 years ago

4 0

Using the discriminant from the quadratic formula (√b²-4ac) since that is the only way to get a solution to not be real (b²-4ac has to be negative), we get
m²-4*3*7=m²-84. Solving for m if m²-84=0, m²=84 and m=+-√84. Therefore, if m is positive, the highest possible value of m is √84. However, we have to find the integer value - with a bit of guess and check, we find that 9²=81 and is the integer you get by rounding down √84 (if we rounded up, 10 would have real solutions since we need the square to be less than 84 to get m²-84 negative). Therefore, 9 is our answer

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If the point (2-3) lies on the graph of the equation 2y=ax-7 then find the value of a <br>

11111nata11111 [884]

Answer:

$\frac{1}{2}$

Step-by-step explanation:

if to substitute the coordinates of the point into the equation:

2*(-3)=2*a-7, then a=1/2

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3 years ago

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Lady bird [3.3K]

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4 years ago

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djverab [1.8K]

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3 years ago

Read 2 more answers

The smallest object visible with your eyes is similar to the width of a piece of hair, which is 1×10−4 meters wide. Using an opt

Sloan [31]

Answer:

B. $5\times10^{2}$

Step-by-step explanation:

We are told that the smallest object visible with our eyes is similar to the width of a piece of hair, which is $1\times 10^{-4}$ meters wide.

Using an optical microscope, we can see items up to $2\times 10^{-7}$ meters wide.

To find the objects we can see with our eyes are how much larger than the objects we can see with an optical microscope, we can set an equation as:

$\frac{\text{The width of the object we can see with our eyes}}{\text{The width of the objects we can see with microscope}}=\frac{1*10^{-4}}{2*10^{-7}}$

Using the exponent rule of quotient $\frac{a^m}{a^n}=a^{m-n}$ we will get,

$\frac{\text{The width of the object we can see with our eyes}}{\text{The width of the objects we can see with microscope}}=\frac{1}{2}*10^{-4-(-7)}$

$\frac{\text{The width of the object we can see with our eyes}}{\text{The width of the objects we can see with microscope}}=0.5*10^{-4+7}$

$\frac{\text{The width of the object we can see with our eyes}}{\text{The width of the objects we can see with microscope}}=0.5*10^{3}$

$\frac{\text{The width of the object we can see with our eyes}}{\text{The width of the objects we can see with microscope}}=0.5*10\times 10^{3-1}$

$\text{The object we can see with our eyes}=5\times10^{2}*\text{The objects we can see with microscope}$

Therefore, the objects we can see with our eyes are $5\times10^{2}$ times larger than the objects we can see with an optical microscope and option B is the correct choice.

3 0

3 years ago

Translate: twice the sum of a number and 9 equals 7 into an equation

Ghella [55]

$equals\ 7\to=7\\sum\ of\ a\ number\ and\ 9\to n+9\ (n-the\ number)\\twice\ the\ sum\ ...\to2(n+9)\\\\Answer:\boxed{2(n+9)=7}\\\\solution:\\\\2(n+9)=7\ \ \ \ |divide\ both\ sides\ by\ 2\\n+9=3.5\ \ \ \ |subtract\ 9\ from\ both\ sides\\\boxed{n=-5.5}$

4 0

3 years ago