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3 years ago

Round 375 to the nearest hundred

Mathematics

2 answers:

Aliun [14]3 years ago

8 0

The nearest hundred would be 400

Paraphin [41]3 years ago

3 0

It's 400. Because:
375 round to ten: 380
380 round to hundred: 400
Rules:
If it's 4 or below, stay still
If it's 5 or above, add one

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The lines shown below are parallel. If the green line has a slope of -3/7 what is the slope of the red line?

anastassius [24]

When any lines are parallel that means they have the same slope so the slope of the red line would also have a slope of -3/7

4 0

2 years ago

What is the solution set of the quadratic inequality 6x2+1<0

Free_Kalibri [48]

Answer:

Solution set of the quadratic equation is, Empty set

Step-by-step explanation:

Given the quadratic equation: $6x^2+1>0$

Subtraction property of equality states that you subtract the same number to both sides of an equation.

Subtract both sides by 1 we get;

$6x^2+1-1>0-1$

Simplify:

$6x^2>-1$

Division property of equality states that you divide the same number to both sides of an equation.

Divide both sides by 6 we get;

$\frac{6x^2}{6} >\frac{-1}{6}$

Simplify:

$x^2>\frac{-1}{6}$

For any x in real number there does not exist any number x which satisfy

$x^2>\frac{-1}{6}$ , therefore, there is no solution for this set of the quadratic inequality or in other word we can say that set of the solution is Empty set.

3 0

3 years ago

Read 2 more answers

If g(x) = 4x^2 - 16 were shifted 5 units to the right and 2 down, what would the new equation be?

AURORKA [14]

Answer:

D. h(x) = 4(x - 5)² - 18

Step-by-step explanation:

4 0

3 years ago

Need help with my homework

Volgvan

Answer:

$\displaystyle y=\frac{16-9x^3}{2x^3 - 3}$

$\displaystyle y=-\frac{56}{13}$

Step-by-step explanation:

<u>Equation Solving</u>

We are given the equation:

$\displaystyle x=\sqrt[3]{\frac{3y+16}{2y+9}}$

To make y as a subject, we need to isolate y, that is, leaving it alone in the left side of the equation, and an expression with no y's to the right side.

We have to make it in steps like follows.

Cube both sides:

$\displaystyle x^3=\left(\sqrt[3]{\frac{3y+16}{2y+9}}\right)^3$