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3 years ago

7

The set of relative max/min and absolute max/min are referred to as _______.

1 answer:

PIT_PIT [208]3 years ago

8 0

Answer:

B

Step-by-step explanation:

You might be interested in

Ainat [17]

<span>(x – h)^2 + (y – k)^2 = r<span>^2

this equation is a derivative of the equation of a circle

x^2 + y^2 = r^2
This is from the origin. If we move the in x or y then the radius will change positions in x or y

with h = -3 and k = 1

we can plug in each set of numbers and solve.

we find Z to be on the circle edge!</span></span>

8 0

3 years ago

Compare the values of each 3 in the number 633,248

ipn [44]

633,248.
The difference of each 3 is that they have different place values.
The 3 on the right side is int he 1000s (thousands) place, and the 3 on the left is in the 10,000s (ten thousands) place. :)

The one in the 10,000s place is 10 times larger than the one in the 1,000s place!

6 0

3 years ago

When the sum of \, 528 \, and three times a positive number is subtracted from the square of the number, the result is \, 120. F

aleksandr82 [10.1K]

Let $x$ be the unknown number. So, three times that number means $3x$ , and the square of the number is $x^2$

We have to sum 528 and three times the number, so we have $528+3x$

Then, we have to subtract this number from $x^2$ , so we have

$x^2-(3x+528)$

The result is 120, so the equation is

$x^2 - 3x - 528 = 120 \iff x^2 - 3x - 648 = 0$

This is a quadratic equation, i.e. an equation like $ax^2+bx+c=0$ . These equation can be solved - assuming they have a solution - with the following formula

$x_{1,2} = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

If you plug the values from your equation, you have

$x_{1,2} = \dfrac{3\pm\sqrt{9-4\cdot(-648)}}{2} = \dfrac{3\pm\sqrt{9+2592}}{2} = \dfrac{3\pm\sqrt{2601}}{2} = \dfrac{3\pm51}{2}$

So, the two solutions would be

$x = \dfrac{3+51}{2} = \dfrac{54}{2} = 27$

$x = \dfrac{3-51}{2} = \dfrac{-48}{2} = -24$

But we know that x is positive, so we only accept the solution $x = 27$

6 0

3 years ago

1. Vinay applied the distributive property to the equation -5(m - 2) - 25. His equation then became --5m + 10 = -125. Did he app

Ahat [919]

Answer:

C. No. Vinay also multiplied the right-hand side of the equation by --5, which is incorrect.

Step-by-step explanation:

<u>Proper application of the property:</u>

- 5(m - 2) - 25 = -5m - 5(-2) - 25 = -5m + 10 - 25 = -5m - 15

Vinay multiplied 5 and 25 as well, which is incorrect.

Correct answer choice is C.

8 0

3 years ago

Ray has found that his new car gets 31 miles per gallon on the highwayand 26 miles per gallon in the city. He recently drove 285

Mkey [24]

Answer:

$Highway = 155\ miles$

$City = 130\ miles$

Step-by-step explanation:

Given

$Highway = 31mi/gallon$

$City = 26mi/gallon$

$Total\ Miles = 285$

$Total\ Gallons = 10$

Required

Determine the number of miles driven on the highway and on the city

Represent the gallons used on highway with h and on city with c.

So, we have:

$c + h = 10$ ---- gallons used

and

$31h + 26c = 285$ --- distance travelled

In the first equation, make c the subject

$c = 10 - h$

Substitute 10 - h for c in the second equation

$31h + 26c = 285$

$31h + 26(10 - h) = 285$

Open bracket

$31h + 260 - 26h = 285$

Collect like terms

$31h - 26h = 285 - 260$

$5h = 25$

Make h the subject

$h = \frac{25}{5}$

$h = 5$

Substitute 5 for h in $c = 10 - h$

$c = 10 - 5$

$c = 5$

If on the highway, he travels 31 miles per gallon, then his distance on the highway is:

$Highway = 31 * 5$

$Highway = 155\ miles$

If in the highway, he travels 26 miles per gallon, then his distance on the highway is:

$City = 26 * 5$

$City = 130\ miles$

8 0

3 years ago