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He area of circle Z is 64 ft2. What is the value of r?

dolphi86 [110]

Answer:

8√π/π

Step-by-step explanation:

A = 64 = π·r²

r² = 64/π

= √64/√π

= 8/√π

= 8√π/π

3 0

3 years ago

Help I need you right nowwwwwwwwwwww

earnstyle [38]

Answer:

6/8

just add the fractions together after making the bottom the same

4 0

3 years ago

Read 2 more answers

C) Five tens seven tenths d) 2 hundreds six ones 3 tenths Change it into decimal☝️

monitta

Answer:

c) 50.7

d) 206.3

Step-by-step explanation:

c) 50 + 0.7 = 50.7

d) 200 + 6 + 0.3 = 206.3

7 0

4 years ago

What is the distance between (6, 1) and (1, -9)?

AVprozaik [17]

Answer: √15 units

Step-by-step explanation:

Let (6,1) be (x^1,y^1) and (1,-9) be (x^2,y^2) .

As we know ,

Distance(D) = √(x^1-x^2) +(y^1-y^2)

Now,

D= √(x^1-x^2) +(y^1-y^2)

= √(6-1) +(1+9)

= √5+10

= √15 units

: Therefore the distance between (6,1) and (1,-9) is √15 units.

4 0

3 years ago

I tell you these facts about a mystery number, $c$: $\bullet$ $1.5 < c < 2$ $\bullet$ $c$ can be written as a fraction wit

makkiz [27]

Answer:

Possible answer: $\displaystyle c = \frac{16}{10} = \frac{8}{5} = 1.6$ .

Step-by-step explanation:

Rewrite the bounds of $c$ as fractions:

The simplest fraction for $1.5$ is $\displaystyle \frac{3}{2}$ . Write the upper bound $2$ as a fraction with the same denominator:

$\displaystyle 2 = 2 \times 1 = 2 \times \frac{2}{2} = \frac{4}{2}$ .

Hence the range for $c$ would be:

$\displaystyle \frac{3}{2} < c < \frac{4}{2}$ .

If the denominator of $c$ is also $2$ , then the range for its numerator (call it $p$ ) would be $3 < p < 4$ . Apparently, no whole number could fit into this interval. The reason is that the interval is open, and the difference between the bounds is less than $2$ .

To solve this problem, consider scaling up the denominator. To make sure that the numerator of the bounds are still whole numbers, multiply both the numerator and the denominator by a whole number (for example, 2.)

$\displaystyle \frac{3}{2} = \frac{2 \times 3}{2 \times 2} = \frac{6}{4}$ .

$\displaystyle \frac{4}{2} = \frac{2\times 4}{2 \times 2} = \frac{8}{4}$ .

At this point, the difference between the numerators is now $2$ . That allows a number ( $7$ in this case) to fit between the bounds. However, $\displaystyle \frac{1}{c} = \frac{4}{7}$ can't be written as finite decimals.

Try multiplying the numerator and the denominator by a different number.

$\displaystyle \frac{3}{2} = \frac{3 \times 3}{3 \times 2} = \frac{9}{6}$ .

$\displaystyle \frac{4}{2} = \frac{3\times 4}{3 \times 2} = \frac{12}{6}$ .

$\displaystyle \frac{3}{2} = \frac{4 \times 3}{4 \times 2} = \frac{12}{8}$ .

$\displaystyle \frac{4}{2} = \frac{4\times 4}{4 \times 2} = \frac{16}{8}$ .

$\displaystyle \frac{3}{2} = \frac{5 \times 3}{5 \times 2} = \frac{15}{10}$ .

$\displaystyle \frac{4}{2} = \frac{5\times 4}{5 \times 2} = \frac{20}{10}$ .

It is important to note that some expressions for $c$ can be simplified. For example, $\displaystyle \frac{16}{10} = \frac{2 \times 8}{2 \times 5} = \frac{8}{5}$ because of the common factor $2$ .

Apparently $\displaystyle c = \frac{16}{10} = \frac{8}{5}$ works. $c = 1.6$ while $\displaystyle \frac{1}{c} = \frac{5}{8} = 0.625$ .

8 0

4 years ago

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