What's is the best estimate of 21% of 387

2 answers:

7 0

387x 21%= 81.27 (estimate is 81.00)

x/387 = 21/100 (cross multiple)

387x21= 8127 and 100x x= 100x (then divide each product by 100)

8127/100 = 81.27 100x/100= x

81.27=x

kompoz [17]3 years ago

4 0

21% of 387 is 81.27 Hope it helps

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What's the answer to this

White raven [17]

Step-by-step explanation:

the angle CDB is the supplementary angle to 9x+10 (angle ADC).

that means they stand together for 180°. as all angles around a single point on one side of a line are together always 180°.

and then, the sum of all angles in a triangle is always 180°.

so,

CDB + (4x + 10) + 50 = 180

CDB = 180 - (9x + 10) = 180 - 9x - 19 = 170 - 9x

therefore,

(170 - 9x) + (4x + 10) + 50 = 180

170 - 9x + 4x + 10 + 50 = 180

-5x = -50

5x = 50

x = 10

angle CDB = 170 - 9x = 170 - 90 = 80°

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valina [46]

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Step-by-step explanation:

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3 years ago

Let log_(b)A=3; log_(b)C=2; log_(b)D=5 what is the value of log_(b)((A^(5)C^(2))/(D^(6)))

Colt1911 [192]

Answer:

-11

Step-by-step explanation:

Given

$\log_bA=3\\ \\\log_bC=2\\ \\\log_bD=5$

Use properties:

$\log_b(A\cdot C)=\log_bA+\log_bC\\ \\\log_b\dfrac{A}{C}=\log_bA-\log_bC\\ \\\log_bA^k=k\log_bA$

Thus,

$\log_b\dfrac{A^5\cdot C^2}{D^6}\\ \\=\log_b(A^5\cdot C^2)-\log_bD^6\\ \\=\log_bA^5+\log_bC^2-\log_bD^6\\ \\=5\log_bA+2\log_bC-6\log_bD\\ \\=5\cdot 3+2\cdot 2-6\cdot 5\\ \\=15+4-30\\ \\=-11$