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3 years ago

Determine whether 270.5m . 30.5m is equivalent to each of the following expressions

Mathematics

1 answer:

Leno4ka [110]3 years ago

8 0

Answer:

The answer is 3m^2

Step-by-step explanation:

Work is shown

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Please help thank you

Eduardwww [97]

Answer:

its (C) or (B) but my answer would be (B)

Step-by-step explanation:

3 0

3 years ago

A tank currently holds 5 gallons of water. It is being filled at the rate of 1/4 gallon per minute. How long will it take for th

Verdich [7]

So we know the tank already has 5 gallons and we need 9. This means we need 4 more gallons in order to fill up the tank.

If 1/4 gallon is being filled per minute we need to find out how many 1/4 gallons makes up 4 gallons in total.

4 of the 1/4 gallons makes up 1 whole gallon. So since we need 4 whole gallons we would need 16 of the 1/4 gallons.

This means that it will take 16 minutes for the tank to hold 9 gallons of water.

6 0

3 years ago

Read 2 more answers

Which triangle always has only two sides the same length and one angle that measures 90°? A. acute scalene B. right isosceles C.

Otrada [13]

Right isosceles, which has two sides the same length and one angle that measures 90 degrees.

5 0

4 years ago

A cash box contains $74 made up of quarters, half-dollars, and one-dollar

Strike441 [17]

Answer:

25 one-dollar coins, 16 half-dollar coins, and 164 quarters

Step-by-step explanation:

First, set up equations based on the information given:

$0.25q+0.50h+1.00d=74$

$\displaystyle{h=\frac{3}{5}d+1}$

$q=4(d+h)$

Then, substitute q in the first equation with the expression from the third equation:

$0.25[4(d+h)]+0.50h+1.00d=74\\1d+1h+0.50h+1.00d=74\\2d+1.5h=74$

Next, substitute h in that equation with the expression from the second equation:

$\displaystyle{2d+1.5(\frac{3}{5}d+1)=74}$

$2d+0.9d+1.5=74\\2.9d+1.5=74$

Solve for d, the number of one-dollar coins:

$2.9d+1.5=74\\2.9d=72.5\\d=25$

Substitute 25 for d in the second equation to find h, the number of half-dollar coins:

$\displaystyle{h=\frac{3}{5}d+1}$

$\displaystyle{h=\frac{3}{5}(25)+1}$

$h=15+1\\h=16$

Substitute 25 for d and 16 for h in the third equation to find q, the number of quarters:

$q=4(d+h)\\q=4(25+16)\\q=4(41)\\q=164$

Then, verify that the coins total $74:

$0.25(164)+0.50(16)+1.00(25)=74\\41+8+25=74\\74=74\\\text{Check.}$

Next, verify that the number of half-dollar coins is one more than three-fifths of the number of one-dollar coins:

$\displaystyle{h=\frac{3}{5}d+1}$

$\displaystyle{16=\frac{3}{5}(25)+1}$

$16 = 15 + 1\\16 = 16\\\text{Check.}$

Finally, verify that the number of quarters is four times the number one-dollar and half-dollar coins together:

$q=4(d+h)\\164=4(25+16)\\164=4(41)\\164=164\\\text{Check.}$

6 0

3 years ago

Find f(a), f(a+h), and 71. f(x) = 7x - 3 f(a+h)-f(a) h if h = 0. 72. f(x) = 5x²

Leni [432]

Answer:

$71. \ \ \ f(a) \ = \ 7a \ - \ 3; \ f(a+h) \ = \ 7a \ + \ 7h \ - \ 3; \ \displaystyle\frac{f(a+h) \ - \ f(a)}{h} \ = \ 7$

$72. \ \ \ f(a) \ = \ 5a^{2}; \ f(a+h) \ = \ {5a}^{2} \ + \ 10ah \ + \ {5h}^{2}; \ \displaystyle\frac{f(a+h) \ - \ f(a)}{h} \ = \ 10a \ + \ 5h$

Step-by-step explanation:

In single-variable calculus, the difference quotient is the expression

$\displaystyle\frac{f(x+h) \ - \ f(x)}{h}$ ,

which its name comes from the fact that it is the quotient of the difference of the evaluated values of the function by the difference of its corresponding input values (as shown in the figure below).

This expression looks similar to the method of evaluating the slope of a line. Indeed, the difference quotient provides the slope of a secant line (in blue) that passes through two coordinate points on a curve.

$m \ \ = \ \ \displaystyle\frac{\Delta y}{\Delta x} \ \ = \ \ \displaystyle\frac{rise}{run}$ .

Similarly, the difference quotient is a measure of the average rate of change of the function over an interval. When the limit of the difference quotient is taken as h approaches 0 gives the instantaneous rate of change (rate of change in an instant) or the derivative of the function.

Therefore,

$71. \ \ \ \ \ \displaystyle\frac{f(a \ + \ h) \ - \ f(a)}{h} \ \ = \ \ \displaystyle\frac{(7a \ + \ 7h \ - \ 3) \ - \ (7a \ - \ 3)}{h} \\ \\ \-\hspace{4.25cm} = \ \ \displaystyle\frac{7h}{h} \\ \\ \-\hspace{4.25cm} = \ \ 7$

$72. \ \ \ \ \ \displaystyle\frac{f(a \ + \ h) \ - \ f(a)}{h} \ \ = \ \ \displaystyle\frac{{5(a \ + \ h)}^{2} \ - \ {5(a)}^{2}}{h} \\ \\ \-\hspace{4.25cm} = \ \ \displaystyle\frac{{5a}^{2} \ + \ 10ah \ + \ {5h}^{2} \ - \ {5a}^{2}}{h} \\ \\ \-\hspace{4.25cm} = \ \ \displaystyle\frac{h(10a \ + \ 5h)}{h} \\ \\ \-\hspace{4.25cm} = \ \ 10a \ + \ 5h$

4 0

2 years ago