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3 years ago

Here is an expression: 3+ t2 1 Evaluate the expression when t is 2

Mathematics

1 answer:

Zanzabum3 years ago

4 0

Step-by-step explanation:

it would be 45 if you add it as 21 but you didn't put another symbol hope I can help :)

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Step-by-step explanation: Money loses value over time as the rate of inflation increases. In this scenario, when you have some money to spare, the ideal is to seek some kind of investment that earns interest.

Saving is one such option. The money in savings earns interest and is highly liquid, meaning it can be used at any time. In addition, saving is a very...

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3 years ago

Ryan earned x number of dollars babysitting his little brother in July. His older

aliya0001 [1]

Answer:

Ryan earned $52 Mike earned $109 and Stormy earned $94

Step-by-step explanation:

x+2×+5+2x-10=255

5x-5=255

5x=260

x=52

6 0

3 years ago

A circle has a radius of 7 cm. Workout the area of the circle. Give your answer correct to three significant figures.

anastassius [24]

Answer:

πr²=22/7×49

=22×7

=154

hope it helps

Mark me brainliest

5 0

3 years ago

Find the solution of Equation: (3x-2)/(x+3)-1=(3x-3)/(x+1)-2

saveliy_v [14]

Answer:

Solution tends to infinity

Step-by-step explanation:

Given the expression

(3x-2)/(x+3)-1=(3x-3)/(x+1)-2

This can also be expressed as;

(3x-2)/x+2 = 3x-3/x-1

3x-2/x+2 = 3(x-1)/x-1

3x-2/x+2 = 3

Cross multiply

3x-2 = 3(x+2)

3x-2 =3x+6

Collect like terms

3x-3x =6+3

0x = 9

x = 9/0

x = infinity

Hence the expression had no solution. It tends to infinity

3 0

3 years ago

Sam entered three functions into a graphing calculator, y1 = x + 2, y2 = x2 + 2, and y3 = 2x. A portion of the table created by

V125BC [204]

For this case we have the following functions:
$y1 = x + 2

y2 = x ^ 2 + 2

y3 = 2 ^ x$

When x = 0 we have:
For y1:
$y1 = 0 + 2

y1 = 2$
For y2:
$y2 = 0 ^ 2 + 2

y2 = 0 + 2

y2 = 2$
Therefore, we have to:
$y1 = y2
$

When x = 5 we have:
For y2:
$y2 = 5 ^ 2 + 2

y2 = 25 + 2

y2 = 27$
For y3:
$y3 = 2 ^ 5

y3 = 32$
Therefore, we have to:
$y2 \ \textless \ y3
$

When x = -1 we have:
For y1:
$y1 = -1 + 2

y1 = 1$
For y2:
$y2 = (-1) ^ 2 + 2

y2 = 1 + 2

y2 = 3
$
For y3:
$y3 = 2 ^ {-1}

y3 = 1/2

y3 = 0.5$
Therefore, we have to:
$y3 \ \textless \ y1 \ \textless \ y2
$

Answer:
When x = 0, y1 = y2
When x = 5, y2 <y3
When x = -1, y3 <y1 <y2

4 0

3 years ago

Here is an expression: 3+ t2 1 Evaluate the expression when t is 2​

Here is an expression: 3+ t2 1 Evaluate the expression when t is 2