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3 years ago

Alex wants to put a fence around a square garden. The garden has an area of 784 ft2.

Mathematics

2 answers:

sweet [91]3 years ago

8 0

Your answer is 392 ft

Irina18 [472]3 years ago

7 0

P = 4s
P = 4(28)
P = 112 ft

The perimeter of the garden is 112 feet.

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Y = log2 (x – 4)<br><br> How do I solve this

Dominik [7]

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7 0

3 years ago

The table shows the outputs, y, for different inputs, x: Input (x) 6 12 15 25 Output (y) 14 15 15 16 Part A: Do the data in this

cluponka [151]

Answer:

Part A: No it is not a function

Part B: The equation

Part C: x = 3

Step-by-step explanation:

Part A - The table:

Input (x) Output (y)

6 14

12 15

15 15

25 16

This is a function because it has no repeating inputs values which have alternate outputs. Each input is assigned exactly one output.

Part B - The relation y = 7x - 15 has the value y = 27 when x = 6. You can find it by substituting into the equation.

y = 7(6) -15

y = 42 - 15

y = 27

The value of the relation in the table when x = 6 is y = 14. The equation has the greater value.

Part C: To find when y = 6, set it equal to 6 and solve for x.

6 = 7x - 15

21 = 7x

3 = x

5 0

3 years ago

Joe began to increase the speed of his car at 2:00 P.M. Since that time,the speed of Joes car has been steadily increasing by 1

Olegator [25]

Answer:

7 mins

Step-by-step explanation:

Current speed of Joes Car = 65.5 mph

We have to find the time interval for which the car exceeded the speed limit of 55 mph.

While, we are given that the speed of the car was constantly increasing, hence the speed over all increased from the limit of 55 mph = 65.50-55.00 = 10.50 mph

We are also given that Joes car was increasing speed at a constant rate of 1.50 mph for every passing minute. Hence

1.50 mph is increased in 1 minute

1 mph will be increase in $\frac{1}{1.5}$ minutes

Hence

10.50 mph will be increased in $\frac{1}{1.5} \times 10.50$ minutes

$= \frac{10.5}{1.5}$

$=7$

Hence joes car was exceeding the limit of 55 mph for 7 minutes.

8 0

3 years ago

A class of 33 students elected a class treasurer. there were 111votes for candidate a and 15 votes for candidate

Advocard [28]

Since 111 is greater than 33, I will assume that you meant 11 votes for a candidate. Also, assuming that each student can vote only once, the answer would be

11/33 x 100=33.333333...

About 33.333 of the voters voted for candidate a.

3 0

3 years ago

Anyone know how to solve?

sergejj [24]

Answer:

was there more

Step-by-step explanation:

(i will edit answer)

3 0

3 years ago