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2 years ago

3 1/2-2/3 help plz its fraction and Idgaf to understand xd

Mathematics

2 answers:

Hatshy [7]2 years ago

8 0

Answer:

2 5/6

Step-by-step explanation:

WE CHANGE 3 1/2 INTO IMPROPER FRACTION WHICH IS 7/2

7/2-2/3=2 5/6

Kisachek [45]2 years ago

5 0

Answer:

Step-by-step explanation:

3 1/3 - 2/3

put it to improper fractions and then get answer as 2/3

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Tommy buys a football for $20 plus 8% tax John buys a helmet for $25 plus 8% tax what is the difference in the amount Tommy and

max2010maxim [7]

Answer:

$5.40

Step-by-step explanation:

First, figure out the cost of the items with tax; for Tommy we have 20*0.08 = 1.6, add that 1.6 to the 20 and his purchase is $21.6, for John we have 25*0.08 = 2, add 2 to his 25 and his purchase is $27. To find the difference we subtract 27-21.6 = 5.4 or $5.40

3 0

2 years ago

What is the sum of the first 70 consecutive odd numbers? Explain.

expeople1 [14]

The sum of the first n odd numbers is n squared! So, the short answer is that the sum of the first 70 odd numbers is 70 squared, i.e. 4900.

Allow me to prove the result: odd numbers come in the form 2n-1, because 2n is always even, and the number immediately before an even number is always odd.

So, if we sum the first N odd numbers, we have

$\displaystyle \sum_{i=1}^N 2i-1 = 2\sum_{i=1}^N i - \sum_{i=1}^N 1$

The first sum is the sum of all integers from 1 to N, which is N(N+1)/2. We want twice this sum, so we have

$\displaystyle 2\sum_{i=1}^N i = 2\cdot\dfrac{N(N+1)}{2}=N(N+1)$

The second sum is simply the sum of N ones:

$\underbrace{1+1+1\ldots+1}_{N\text{ times}}=N$

So, the final result is

$\displaystyle \sum_{i=1}^N 2i-1 = 2\sum_{i=1}^N i - \sum_{i=1}^N 1 = N(N+1)-N = N^2+N-N = N^2$

which ends the proof.

5 0

2 years ago

If x ≠ 2y, then <br><br> eq1a<br><br><br> 2(x-2y)<br><br> 2y-x<br><br> 1<br><br> 0<br><br> -2

lesya692 [45]

The answer is 0 because they’re the same

8 0

2 years ago

PLEASE PLEASE HELP! Find the measure of APB^.

Whitepunk [10]

Answer:

65°

Step-by-step explanation:

Radii CA and CB are perpendicular to tangent lines AT and BT, so

$\angle CAT=\angle CBT=90^{\circ}$

Since angle BAT is equal to 65°, angle CAB has measure

$\angle CAB=90^{\circ}-65^{\circ}=25^{\circ}.$

Consider triangle ACB. This triangle is isosceles, because CA=CB as radii of the circle. Two angles adjacent to the base are congruent, thus

$\angle CBA=\angle CAB=25^{\circ}$

The sum of the measures of all interior angles in triangle is always 180°, so

$\angle CAB+\angle CBA+\angle ACB=180^{\circ}\\ \\25^{\circ}+25^{\circ}+\angle ACB=180^{\circ}\\ \\\angle ACB=130^{\circ}$

Angle ACB is central angle subtended on the minor arc AB, angle APB is inscribed angle subtended on the same minor arc AB. The measure of inscribed angle is half the measure of central angle subtended on the same arc, so

$x=\angle APB=\dfrac{1}{2}\cdot 130^{\circ}=65^{\circ}$