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hammer [34]
3 years ago
15

What is five odd numbers greater than 500

Mathematics
1 answer:
valentinak56 [21]3 years ago
6 0

Step-by-step explanation:

Five odd numbers greater than 500 are 501, 503, 505, 507, 509.

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When solving this equation: −16 + x = −15 we start with adding 16 to both sides. Why do we use addition?
noname [10]

Answer:

We use addition because -16 is negative, and we need to cancel out its value so we add 16 to both sides, leaving 0+x, or x, on the left and combining like terms on the right.

Step-by-step explanation:

5 0
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Pleaseee help me!! ASAP
vichka [17]

Answer:

Yes 4:3

Step-by-step explanation:

To solve this, take the the inside number and subtract in from the outside number. Look at the question very carefully for this part...

Since it says the ratio of the outside number to the inside number, take that number and see how many times it can fit into the outside number. In this case, it is 4, and than for the inside number, it is 3.

Now do the same for the other measurement and see if it has the same ratio, and if it does, that is the answer.

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help me thank you whoever is right gets brainliest A carpenter attaches a piece of plywood with a width of 4 feet and a length o
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Step-by-step explanation:

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3 years ago
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68 more than eight times a number is greater than 7 times the number what numbers satisfy this inequality
JulsSmile [24]

Answer:

x > -68

Step-by-step explanation:

68 + 8x > 7x

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7 0
3 years ago
 Find sin2x, cos2x, and tan2x if sinx=-15/17 and x terminates in quadrant III
Paha777 [63]

Given:

\sin x=-\dfrac{15}{17}

x lies in the III quadrant.

To find:

The values of \sin 2x, \cos 2x, \tan 2x.

Solution:

It is given that x lies in the III quadrant. It means only tan and cot are positive and others  are negative.

We know that,

\sin^2 x+\cos^2 x=1

(-\dfrac{15}{17})^2+\cos^2 x=1

\cos^2 x=1-\dfrac{225}{289}

\cos x=\pm\sqrt{\dfrac{289-225}{289}}

x lies in the III quadrant. So,

\cos x=-\sqrt{\dfrac{64}{289}}

\cos x=-\dfrac{8}{17}

Now,

\sin 2x=2\sin x\cos x

\sin 2x=2\times (-\dfrac{15}{17})\times (-\dfrac{8}{17})

\sin 2x=-\dfrac{240}{289}

We know that,

\cos 2x=1-2\sin^2x

\cos 2x=1-2(-\dfrac{15}{17})^2

\cos 2x=1-2(\dfrac{225}{289})

\cos 2x=\dfrac{289-450}{289}

\cos 2x=-\dfrac{161}{289}

Using the trigonometric ratios, we get

\tan 2x=\dfrac{\sin 2x}{\cos 2x}

\tan 2x=\dfrac{-\dfrac{240}{289}}{-\dfrac{161}{289}}

\tan 2x=\dfrac{240}{161}

Hence, the required values are \sin 2x=-\dfrac{240}{289},\cos 2x=-\dfrac{161}{289},\tan 2x=\dfrac{240}{161}.

6 0
3 years ago
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