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rjkz [21]
3 years ago
12

Nightguy6789 and rebeccamccain11 what is 17x 15??

Mathematics
2 answers:
jeyben [28]3 years ago
6 0

Answer:

255

Step-by-step explanation:

17x15=255

maxonik [38]3 years ago
5 0

Answer:

255

Step-by-step explanation:

17 x 15 = 255

Hope this helps

-Amelia

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dalvyx [7]

Answer:

1) 5.8

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7 0
3 years ago
If the sixth term of a sequence is 128 and the common ratio is 2, then what is the first term? 1, 2 or ,4
stiks02 [169]

the  Answer C: 4

Step-by-step explanation:

6 0
3 years ago
A large box of crackers away 20 Oz if James needs 3 pounds of crackers for a party how many boxes of crackers should he buy
Alenkinab [10]

Answer:

James will need to buy 3 boxes of crackers for the party.

Step-by-step explanation:

remember: 16oz=1 pound

If James needs 3 pounds of crackers, and each box is 20oz, that means each box is a little over 1 pound. (1.25 pounds to be exact). So, we know that he needs to buy 3 boxes of crackers. This is because 20×3=60 ounces which is 3.75 pounds. Meaning, 3 boxes is the correct answer since we can't cut a box in half.

4 0
2 years ago
The beach ball we played with has a density of 10 grams per cubic centimeter
Wittaler [7]
Answer: does not make sense.

The statement " t<span>he beach ball we played with has a density of 10 grams per cubic centimeter" does not make sense.

 Reasoning.

Inflated beach balls have a very low density, they are pretty much only air, plus the weight of the ball per se. Certainly that density is much lower than the density of water.

Inflated beach balls fall slowly in the air because their density is very close to the density of air and float on water because their density is less than the density of water.

Being the density of pure water 1 g/cm^3 you can predict, without any calculation, that the density of the beach ball is less than that and never 10 g/cm^3 which is 10 times the density of pure water.
</span>
3 0
3 years ago
Please help me!!!!!!!!!!!!!​
monitta

Answer:  see proof below

<u>Step-by-step explanation:</u>

Use the following Half-Angle Identities:    tan (A/2) = (sinA)/(1 + cosA)

                                                                     cot (A/2) = (sinA)/(1 - cosA)

Use the Pythagorean Identity: cos²A + sin²B = 1

Use Unit Circle to evaluate: cos 45° = sin 45° = \frac{\sqrt2}{2}

<u>Proof LHS → RHS</u>

Given:                       cot\ (22\frac{1}{2})^o-tan\ (22\frac{1}{2})^o

Rewrite Fraction:     cot\ (\frac{45}{2})^o-tan\ (\frac{45}{2})^o

Half-Angle Identity:   \dfrac{sin(45)^o}{1-cos(45)^o}-\dfrac{sin(45)^o}{1+cos(45)^o}

Substitute:                  \dfrac{\frac{\sqrt2}{2}}{1-\frac{\sqrt2}{2}}-\dfrac{\frac{\sqrt2}{2}}{1+\frac{\sqrt2}{2}}

Simplify:                      \dfrac{\frac{\sqrt2}{2}}{\frac{2-\sqrt2}{2}}-\dfrac{\frac{\sqrt2}{2}}{\frac{2+\sqrt2}{2}}

                               =\dfrac{\sqrt2}{2-\sqrt2}-\dfrac{\sqrt2}{2+\sqrt2}

                               =\dfrac{\sqrt2}{2-\sqrt2}\bigg(\dfrac{2+\sqrt2}{2+\sqrt2}\bigg)-\dfrac{\sqrt2}{2+\sqrt2}\bigg(\dfrac{2-\sqrt2}{2-\sqrt2}\bigg)

                               =\dfrac{2\sqrt2+2}{4-2}-\dfrac{2\sqrt2-2}{4-2}

                               =\dfrac{4}{2}

                               = 2

LHS = RHS:  2 = 2  \checkmark

7 0
4 years ago
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