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kogti [31]
3 years ago
15

Whoever helps me I will mark brainliest.

Mathematics
1 answer:
True [87]3 years ago
3 0

Answer:

Option no. D is right. Because the congruence presented is more congruent

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The population of a place increased to 54,000 in 2003 at a rate of 5% per annum. what would
svlad2 [7]

Answer:

The population in 2005 is 59,535 people.

Step-by-step explanation:

Annum means annually or yearly. We assume the rate has been consistent. To calculate percentage, take the percent in decimal form. 5% is 0.05 (like 17% is 0.17, and 50% 0.5). 5% is actually 5 hundredths of a number. Multiply this by the number you want the percentage of. 54,000*0.05=2,700. This is how much the population grows from 2003 to 2004.

There is a very common mistake people make after this, which is simply doubling that number and adding to the population. This is incorrect because if it grows 5% yearly, the yearly population CHANGES. The question would change if it were 2004 to 2005. Ask your teacher for clarification if this sounds wrong (if it is, do what I said above, the answer would be 59,400 people).

So, the population in 2004 is 54,000+2,700=56,700. Now, find 5% of the population in 2004 using the same method as before. 56,700*0.05=2,835. Add the growth to the pervious population. 56,700+2,835=59,535.

The population in 2005 is 59,535 people.

7 0
2 years ago
Evaluate the integral. W (x2 y2) dx dy dz; W is the pyramid with top vertex at (0, 0, 1) and base vertices at (0, 0, 0), (1, 0,
In-s [12.5K]

Answer:

\mathbf{\iiint_W (x^2+y^2) \ dx \ dy \ dz = \dfrac{2}{15}}

Step-by-step explanation:

Given that:

\iiint_W (x^2+y^2) \ dx \ dy \ dz

where;

the top vertex = (0,0,1) and the  base vertices at (0, 0, 0), (1, 0, 0), (0, 1, 0), and (1, 1, 0)

As such , the region of the bounds of the pyramid is: (0 ≤ x ≤ 1-z, 0 ≤ y ≤ 1-z, 0 ≤ z ≤ 1)

\iiint_W (x^2+y^2) \ dx \ dy \ dz = \int ^1_0 \int ^{1-z}_0 \int ^{1-z}_0 (x^2+y^2) \ dx \ dy \  dz

\iiint_W (x^2+y^2) \ dx \ dy \ dz = \int ^1_0 \int ^{1-z}_0 ( \dfrac{(1-z)^3}{3}+ (1-z)y^2) dy \ dz

\iiint_W (x^2+y^2) \ dx \ dy \ dz = \int ^1_0  \ dz \  ( \dfrac{(1-z)^3}{3} \ y + \dfrac {(1-z)y^3)}{3}] ^{1-x}_{0}

\iiint_W (x^2+y^2) \ dx \ dy \ dz = \int ^1_0  \ dz \  ( \dfrac{(1-z)^4}{3}+ \dfrac{(1-z)^4}{3}) \ dz

\iiint_W (x^2+y^2) \ dx \ dy \ dz =\dfrac{2}{3} \int^1_0 (1-z)^4 \ dz

\iiint_W (x^2+y^2) \ dx \ dy \ dz =- \dfrac{2}{15}(1-z)^5|^1_0

\mathbf{\iiint_W (x^2+y^2) \ dx \ dy \ dz = \dfrac{2}{15}}

7 0
3 years ago
A game room has a floor that is 400 feet by 90 feet. A scale drawing of the floor on grid paper uses a scale of 1 unit:5 feet. W
Allisa [31]
First 1/5 is equivalent to 20% so to get 20% of both of them multiply by .2. so 400×.2=80 and 90×.2=18
4 0
3 years ago
In the triangles, GK PN and HG 2MP.
dalvyx [7]

Answer:

G is smaller than P. Visually comparing G to P you are doing a rough comparison to a 90° (or right) angle. Angle G is a little bit tilted to the right making it just under 90°

6 0
3 years ago
What is the value of the expression
Gelneren [198K]

Answer:32

Step-by-step explanation: First you do 4×4×4×

=64 then you divide by 2.. 64÷2=32

8 0
3 years ago
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