Enter the equation of the circle with the given center and radius.

Help me in the first part thanks I will give brainlist

vivado [14]

He will need exactly 24 stickers. To get this number, you must multiply both sides. :)

3 0

2 years ago

Read 2 more answers

I neeed helpppppppppppppppp

Stolb23 [73]

✧･ﾟ: *✧･ﾟ:* *:･ﾟ✧*:･ﾟ✧

Hello!

✧･ﾟ: *✧･ﾟ:* *:･ﾟ✧*:･ﾟ✧

❖ Jeff will earn $267.20 in a 32 hour week.

Divide to find how much he gets paid per hour:

$334 ÷ 40 = $8.35

Multiply the cost per hour by 32 hrs to get the total amount he'll get paid:

$8.35 × 32 = $267.20

~ ʜᴏᴘᴇ ᴛʜɪꜱ ʜᴇʟᴘꜱ! :) ♡

~ ᴄʟᴏᴜᴛᴀɴꜱᴡᴇʀꜱ

5 0

2 years ago

Graph the functions and approximate an x-value in which the exponential function exceeds the polynomial function. y = 4x y = 7x2

nordsb [41]

Answer:

The Exponential function is

$y=4^x$

And the polynomial function is

$y=7 x^2 +4 x -2$

And, we have to find the value of x for which, exponential function exceeds the polynomial function which can be written as

$4^x> 7 x^2 +4 x -2$

1. When , x= -1

LHS

$4^{-1}=\frac{1}{4}=0.25$

RHS

$=7*(-1)^2 +4 *(-1)-2\\\\=7-4-2=1$

2. When , x=0

L HS

$4^0=1$

RHS

$7*0+4*0-2= -2$

3. When ,x= 0.5

L HS

$4^{0.5}=2$

RHS

$=7*0.25 +4*0.5 -2\\\\= 1.75+2-2\\\\=1.75$

4. When , x=2

LHS

$4^{2}=16$

RHS

$=7*4^2+4*4-2\\\\=112+16-2\\\\=126$

The Minimum value for which exponential function exceeds the polynomial function is , x= 0.5

But,there is other value for which exponential function exceeds the polynomial function is , x=2.

5 0

2 years ago

Which is the simplified form of the expression ((p squared) (q Superscript 5 Baseline)) Superscript negative 4 Baseline times ((

Alexus [3.1K]

Answer:

$q^{-30}$

Step-by-step explanation:

Given the expression:

$((p^2) (q ^5)) ^{-4} X ((p ^{-4}) (q^5)) ^{-2}$

First, we open the outer brackets

$=(p^{2X-4}) (q ^{5X-4})X (p ^{-4X-2}) (q^{5X-2})\\\\=(p^{-8}) (q ^{-20})X (p ^{8}) (q^{-10})\\\\$Next, we can collect like terms\\\\=p^{-8}Xp ^{8}Xq ^{-20}Xq ^{-10}\\\\$Applying addition law of indices\\\\=p^{-8+8}Xq^{-20+(-10)}\\\\=p^0 X q^{-30}\\\\=1 X q^{-30}\\\\=q^{-30}$