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Irina-Kira [14]
3 years ago
7

Find the number that should be added to the expression to make it a perfect square. k^2-5k

Mathematics
1 answer:
yuradex [85]3 years ago
6 0
(K-5/2)^2-25/4 is your answer
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How do you solve the substitution of y=5x-3 and y=-6x-3
svlad2 [7]
5x-3=y      6x-3=y
x=1            x=1
51-3=y      61-3=y
51=5         61=6
5-3=y        6-3=y       
2               3
y=2           y=3               


8 0
3 years ago
Pls help I’ll brainlest
svet-max [94.6K]

Answer:

Choice B

Step-by-step explanation:

7.1×10⁵ = 710,000

8 0
2 years ago
Read 2 more answers
Can someone help with step by step
MrMuchimi

Answer:

See explanation

Step-by-step explanation:

In\: \triangle ABC \:\&\: \triangle EDC\\\angle ABC \cong \angle EDC... (Right \: \angle s) \\\angle ACB \cong \angle ECD... (Vertical \: \angle s) \\\therefore \triangle ABC \sim \triangle EDC... (By\: AA\: Postulate) \\

3 0
3 years ago
Perhatikan pola bilangan berikut:2,-6,18,-54,162,...​
andrezito [222]

Answer:

no comprehendo

Step-by-step explanation:

8 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\\

= \int\limits343 cos^{2}  \theta \, d\theta

8 0
3 years ago
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