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3 years ago

Need help with 11 please

Mathematics

1 answer:

salantis [7]3 years ago

4 0

Yes I have yes I have enough because 2000 and 100 equals an ounce of the answers and you would get 2 + 15 + + 12. 5 6 + 3

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Bernice has a total of $0.25 she has some nickels and pennies how many different combinations are nickels and pennies could bian

Natasha_Volkova [10]

There is 4 possible combinations you could make.

4 0

3 years ago

Temp of 20* C will drop by 4*c

svp [43]

Answer:

16* C

Step-by-step explanation:

This is simply 20 - 4

---> 20 - 4 = 16

=> Celcius (C)

Thus, 16° C

4 0

3 years ago

Will be giving brainliest to the correct one, no links or account will be reported:)

aleksandr82 [10.1K]

Answer:

x = 38

∠P = 26º

Step-by-step explanation:

The sum of the angle in a triangle is 180

x + (x + 78) + (x - 12) = 180

Combine like terms

3x + 66 = 180

Subtract 66 from both sides

3x = 114

Divide both sides by 3

x = 38

-----------------------

∠P = x - 12

∠P = 38 - 12

∠P = 26º

7 0

3 years ago

Read 2 more answers

Timothy built a base for a circular tabletop. The base can support a tabletop with a radius of at least 6 inches, but not more t

neonofarm [45]

Let

r------> the radius of the circular tabletop

we know that

$r\geq 6\ in$

$r\leq23\ in$

$6\ in \leq r \leq 23\ in$

The area of a circle is equal to

$A=\pi r^{2}$

where

r is the radius of the circle

Part a) What is the smallest possible area of the tabletop that will fit on Timothy’s table base?

we know that

the smallest possible area of the tabletop is for $r=6\ in$

Substitute the value of r in the formula

$A=\pi 6^{2}$

$A=113.09\ in^{2}$

Round to the nearest whole square inch

$A=113\ in^{2}$

therefore

the answer part a) is

the smallest possible area of the tabletop is $113\ in^{2}$

Part b) What is the largest possible area of the tabletop that will fit on Timothy’s table base?

we know that

the largest possible area of the tabletop is for $r=23\ in$

Substitute the value of r in the formula

$A=\pi 23^{2}$

$A=1,661.90\ in^{2}$

Round to the nearest whole square inch

$A=1,662\ in^{2}$

therefore

the answer part b) is

the largest possible area of the tabletop is $1,662\ in^{2}$

7 0

4 years ago

Read 2 more answers

Find the fourth term of the geometric sequence ,given the first term and common ratio a1=8 and r=4

Rama09 [41]

The equation would be “An= 8(4)^n”. Your answer would be equal to 8 times 4 to the 4th power, which is 2048

5 0

4 years ago