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Ulleksa [173]
3 years ago
13

Hi please help , With solution if it's ok

Mathematics
1 answer:
Snowcat [4.5K]3 years ago
5 0

<u>Answers </u><u>with </u><u>Method</u><u>:-</u>

1) Multiply the length value by 100,000.

2.5 \: hm = 2.5 \times 100000

=  >2.5 \: hm = 250000 \: mm

2) For this, divide the length value by 1000.

=  >  \frac{1800}{1000}

=  > 1.8 \: dam

3) For finding the approximate value, just multiply the value of length by 1.609.

=  >6450 \: m = (6450 \times 1.609) \: km

=  > 10380.27 \: km

4) Multiply the given mass value by 100.

=  > 1.2 \: kg = 1.2 \times 100

=  > 1200 \: g

5) Multiply the given value by 10,000.

4.37 \: dag = 4.37 \times 10000

=  > 43700 \: mg

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x is approximately 25.83

Step-by-step explanation:

x + 25 + 5x = 180

6x + 25 = 180

6x = 155

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Step-by-step explanation:

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A line passes through the points (6,-6) and (9,-5). What is it’s equation in point slope form?
BARSIC [14]

Answer:

y=\frac{1}{3}x-8

Step-by-step explanation:

Slope-intercept form of an equation is written as y=mx+b, where m is the slope and b is the y-intercept.

The slope of a line that passes through the points (x_1,\: y_1) and (x_2, \: y_2) is m=\frac{\Delta y}{\Delta x}=\frac{y_2-y_1}{x_2-x_1}. Using the coordinates (6,-6) and (9,-5) as given in the problem, we have slope of this line to be:

m=\frac{-5-(-6)}{9-6}=\frac{1}{3}.

Now using this slope we've found and any point the line passes through, we can find the y-intercept of this equation:

-6=\frac{1}{3}(6)+b, \\ b=-8

Therefore, the equation of this line in slope-intercept form is \fbox{$y=\frac{1}{3}x-8$}.

3 0
3 years ago
The lateral edge PA of a triangular pyramid ABCP is perpendicular to the base and PA = 2sqrt5 in. The base edges AB = 10 in, AC
DIA [1.3K]

Answer:

PQ = √84 = 2√21 in ≈ 9.165 in

Step-by-step explanation:

The base edges AB = 10 in, AC = 17 in, and BC = 21 in

Q ∈ BC and AQ is the altitude of the base.

Let BQ = x in ⇒ CQ = (21 - x) in

ΔAQB a right triangle at Q

∴ AQ² = AB² - BQ² = 10² - x²  ⇒(1)

ΔAQC a right triangle at Q

∴ AQ² = AC² - CQ² = 17² - (21-x)²  ⇒(2)

Equating (1) and (2)

∴  10² - x² = 17² - (21-x)²

∴ 10² - x² = 17² - (21² - 42x + x²)

∴  10² - x² = 17² - 21² + 42x - x²

∴ 10² - 17² + 21² = 42x

∴ 42x = 252

∴ x = 252/42 = 6

Substitute at (1)

∴ AQ² = AB² - BQ² = 10² - x² = 100 - 36 = 64 = 8²

∴ AQ = 8 in

∵ The lateral edge PA of a triangular pyramid ABCP is perpendicular to the base

∴ PA⊥AQ

∴ ΔPAQ is a right triangle at A,

PA = 2sqrt5 in  and AQ = 8 in

∴ PQ (hypotenuse) = √(PA² + AQ²)

∴ PQ = √[(2√5)² + 8²] = √(20+64) = √84 = 2√21 in ≈ 9.165 in

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3 years ago
4. Which expression is equivalent to (18 7• 5 9 ) 3/ 18 exponent 3
ASHA 777 [7]

Answer:

a. or c.

Step-by-step explanation:

.18 21/5 27

18 10/ 5 12

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3 years ago
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