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Mashutka [201]
3 years ago
7

!! 30 PTS !! Hello, I need some help on this Geometry Area question:

Mathematics
1 answer:
ELEN [110]3 years ago
7 0
You have to multiply 54 times 1.2 and you will get 64.8. So, 64.8 is your area. 
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A triangle with vertices at A(0, 0), B(0, 4), and C(6, 0) is dilated to yield a triangle with vertices at A′(0, 0), B′(0, 10), a
Cerrena [4.2K]
K · ( 0, 0 ) = ( 0 , 0 )
k · ( 0 , 4 ) = ( 0 , 10 )
k · ( 6, 0 ) = ( 15, 0 )
k = 15/6 = 10/4 = 2.5
Answer: The scale of dilation is 2.5
6 0
4 years ago
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write a possible point slope form of the equation of the line that passes through the points (4,4) and (3,-2)?​
Mashutka [201]

m=-2-4/3-4=-6/-1=6

y-4=6(x-4)

y=6x-20

6 0
3 years ago
Solve K/2 + 5 - 2k = 6k
pantera1 [17]

Answer:

k = 2/3 is the answer!

Step-by-step explanation:

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4 0
3 years ago
What is the factorization of 729^15+1000?
igomit [66]

Answer:

The factorization of 729x^{15} +1000 is (9x^{5} +10)(81x^{10} -90x^{5} +100)

Step-by-step explanation:

This is a case of factorization by <em>sum and difference of cubes</em>, this type of factorization applies only in binomials of the form (a^{3} +b^{3} ) or (a^{3} -b^{3}). It is easy to recognize because the coefficients of the terms are <u><em>perfect cube numbers</em></u> (which means numbers that have exact cubic root, such as 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, etc.) and the exponents of the letters a and b are multiples of three (such as 3, 6, 9, 12, 15, 18, etc.).

Let's solve the factorization of 729x^{15} +1000 by using the <em>sum and difference of cubes </em>factorization.

1.) We calculate the cubic root of each term in the equation 729x^{15} +1000, and the exponent of the letter x is divided by 3.

\sqrt[3]{729x^{15}} =9x^{5}

1000=10^{3} then \sqrt[3]{10^{3}} =10

So, we got that

729x^{15} +1000=(9x^{5})^{3} + (10)^{3} which has the form of (a^{3} +b^{3} ) which means is a <em>sum of cubes.</em>

<em>Sum of cubes</em>

(a^{3} +b^{3} )=(a+b)(a^{2} -ab+b^{2})

with a= 9x^{5} y b=10

2.) Solving the sum of cubes.

(9x^{5})^{3} + (10)^{3}=(9x^{5} +10)((9x^{5})^{2}-(9x^{5})(10)+10^{2} )

(9x^{5})^{3} + (10)^{3}=(9x^{5} +10)(81x^{10}-90x^{5}+100)

.

8 0
3 years ago
When the smaller of two consecutive integers is added to two times the​ larger, the result is 26. find the integers?
Mashutka [201]
(X)+2(x+1)=26
3x=24
X=8

Y=x+1
Y=8+1
Y=9
The two numbers are 8 and 9
8 0
3 years ago
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