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3 years ago

Order the values from GREATEST to LEAST: -8 |-8| -17 25 0

Mathematics

2 answers:

Dvinal [7]3 years ago

4 0

25 >|-8| >0 >-8 >-17

Marta_Voda [28]3 years ago

3 0

25 |8| 0 -8 -17 from greatest to least

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9. In a rhombus whose side measures 34 and the smaller angle is 42°. Find the length of the larger diagonal, to the nearest tent

Veseljchak [2.6K]

Answer:

63.5

Step-by-step explanation:

The diagonals of a rhombus bisect opposite angles.

The longer diagonal bisects the 42° angle

The larger angle of the rhombus is 180 - 42 = 138

Let x = the length of the longer diagonal

You know two sides and the included angle of a triangle.

Using the law of cosines we get

$x^{2} = 34^{2} + 34^{2} -2(34)(34) cos 138\\ = 1156 + 1156 - 2312 cos 138\\ = 4030.150837...\\x = \sqrt{4030.150837} = 63.48$ $x^{2} = 34^{2} + 34^{2} -2(34)(34)cos 138\\ = 1156 + 1156 - 2312 cos 138\\ = 4030.1508...\\ x = \sqrt{4030.1508} \\ = 63.48$ x^2 = 34^2 + 34^2 - 2(34)(34) cos 138

= 1156 + 1156 - 2312 c0s 138

= 4030.1508...

x = sq rt 4030.1508...

= 63.5

7 0

2 years ago

How to round 750 to the nearest hundred

Rzqust [24]

The way you round is if it is below 5 you round down and if it is above 5 you round up so in the case of 750 it would go to 800 where if it was 749 it would go to 700

7 0

2 years ago

HELP ME PLZ Between 9 P.M. and 6:20 A.M., the water level in a swimming pool decreased by 7/12 . Assuming that the water level

ivann1987 [24]

Answer:

0.06245

Step-by-step explanation:

Decrease 7/12=0.583

from 9 p.m to 6:20 am is 9:20= (9+20/60) = 9.33 hours

0.583/9.33 = 0.06248

round the 0.06248 to 0.06245

3 0

3 years ago

Evaluate the following integral using trigonometric substitution.

wariber [46]

Answer:

Step-by-step explanation:

1. Given the integral function $\int\limits {\sqrt{a^{2} -x^{2} } } \, dx$ , using trigonometric substitution, the substitution that will be most helpful in this case is substituting x as $asin \theta$ i.e $x = a sin\theta$ .

All integrals in the form $\int\limits {\sqrt{a^{2} -x^{2} } } \, dx$ are always evaluated using the substitute given where 'a' is any constant.

From the given integral, $\int\limits {7\sqrt{49-x^{2} } } \, dx = \int\limits {7\sqrt{7^{2} -x^{2} } } \, dx$ where a = 7 in this case.

The substitute will therefore be $x = 7 sin\theta$

2.) Given $x = 7 sin\theta$

$\frac{dx}{d \theta} = 7cos \theta$

cross multiplying

$dx = 7cos\theta d\theta$

3.) Rewriting the given integral using the substiution will result into;

$\int\limits {7\sqrt{49-x^{2} } } \, dx \\= \int\limits {7\sqrt{7^{2} -x^{2} } } \, dx\\= \int\limits {7\sqrt{7^{2} -(7sin\theta)^{2} } } \, dx\\= \int\limits {7\sqrt{7^{2} -49sin^{2}\theta } } \, dx\\= \int\limits {7\sqrt{49(1-sin^{2}\theta)} } } \, dx\\= \int\limits {7\sqrt{49(cos^{2}\theta)} } } \, dx\\since\ dx = 7cos\theta d\theta\\= \int\limits {7\sqrt{49(cos^{2}\theta)} } } \, 7cos\theta d\theta\\= \int\limits {7\{7(cos\theta)} }}} \, 7cos\theta d\theta\\$