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frez [133]
3 years ago
11

Please I need help asap

Mathematics
2 answers:
Korvikt [17]3 years ago
5 0

Answer:

area of a triangle

1/2 base ×height

1/2×15×8.5

=63.75cm ^2

faltersainse [42]3 years ago
3 0

Answer:

63.75cm^2 / square centimeters

Step-by-step explanation:

The area of a triangle is calculated by multiplying 1/2 times the base, times the height.

A = (1/2)(base)(height)

A = (0.5)(15)(8.5)

A =63.75

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Complete the following table by performing the operations to write equivalent systems.
Anvisha [2.4K]

Answer:

See below:

Step-by-step explanation:

Problem 1:

Multiply Equation 1 by 4, keep Equation 2 the same.

x+y=8, multiply each term by 4:

4*x=4x, 4*y=4y, 8*4=32

so, the equivalent system is: 4x+4y=32 and x-y=2

Solve the system of equations:

x-y=2 becomes x=y+2

plug into 4x+4y=32 to solve for y

4(y+2)+4y=32---> 4y+8+4y=32--->8y=24---> y=3

Plug into x-y=2---> x-3=2---> x=5

Problem 1 Answer:

Equivalent system: 4x+4y=32, x-y=2; solution: x=5, y=3

Problem 2:

Keep Equation 1 the same. Add 1 and 2.

To add an equation, add the left sides together, and then the rights.

so: x+y=8 + x-y=2 gives us: 2x=10

solve for x ---> 2x/2=10/2--->x=5

plug x into x+y=8--->5+y=8--->y=3

Problem 2 answer:

Equivalent system: x+y=8, 2x=10; solution:x=5, y=3

Problem 3:

Subtract Equation 2 from 1, and keep 2 the same.

To subtract an equation, subtract the left sides, then the rights. We are subtracting 1<em> from </em>2, so its 2-1.

x-y=2 - x+y=8 gives us: -2y=-6

Solve for y by dividing by -2-->-2y/-2=-6/-2---> y=3

Plug into x-y=2---> x-3=2---> x=5

Problem 3 answer:

Equivalent system: -2y=-6, x-y=2; solution: x=5, y=3

Problem 4:

Multiply the sum of Equation 1 and 2 by a factor of 3. Keep equation 2 the same.

First we add 1 and 2: (we did this earlier) ---> 2x=10 ---> now we multiply it all by 3---> 2x*(3)=10*(3)---> this gives us: 6x=30---> now divide by 6 to solve for x: 6x/6=30/6 gives us: x=5

Now, solve for y by plugging x into equation 2: x-y=2---> 5-y=2--->y=3

Problem 4 answer:

Equivalent system: 6x=30, x-y=2; solution: x=5, y=3

______

Quick Tip: One thing inherent of Equivalent systems is that they have the same set of solutions. Thus, we know the systems are equivalent when they have the same set of solutions for x and y. Moreover, you don't need to solve every time after you attempt to find an equivalent system, instead, just plug in the values found in problem 1 to each new set of equations to test if they are equivalent.

If we find x=5 and y=3 for x+y=8 and x-y=2, then all we have to do is plug them in to 6x=30 and -2y=-6 to see if they are equivalent.

6(5)=30 ---> true

-2(3)=-6 ---> true

8 0
4 years ago
What is the highest common factors and lowest common multiple of 36 and 84 using prime factors​
krok68 [10]

Answer:

The corresponding divisor (12) is the GCF of 36 and 84.

2nd answer

LCM = 252

Step-by-step explanation:

Divide 84 (larger number) by 36 (smaller number). Since the remainder ≠ 0, we will divide the divisor of step 1 (36) by the remainder (12). Repeat this process until the remainder = 0. The corresponding divisor (12) is the GCF of 36 and 84.

2nd answer step

Find the prime factorization of 36 36 = 2 × 2 × 3 × 3 Find the prime factorization of 84 84 = 2 × 2 × 3 × 7 Multiply each factor the greater number of times it occurs in steps i) or ii) above to find the LCM: LCM = 2 × 2 × 3 × 3 × 7 LCM = 252

8 0
2 years ago
Read 2 more answers
Luke is going to reflect point R(6, 10) over the y-axis what are the coordinates for R
creativ13 [48]

Answer:

R - (-6,10)

Flip the x-value when reflecting over y axis, and flip y-value when reflecting over x-axis

8 0
3 years ago
Use scientific notation to estimate the number of hours in one year
Studentka2010 [4]
8.76582 × 10 to the 3rd power, that's how many hours there are in a year using scientific notation.
4 0
4 years ago
Read 2 more answers
Using fermat's little theorem, find the least positive residue of $2^{1000000}$ modulo 17.
torisob [31]
Fermat's little theorem states that
a^p≡a mod p

If we divide both sides by a, then
a^{p-1}≡1 mod p
=>
a^{17-1}≡1 mod 17
a^{16}≡1 mod 17

Rewrite
a^{1000000} mod 17  as
=(a^{16})^{62500} mod 17
and apply Fermat's little theorem
=(1)^{62500} mod 17
=>
=(1) mod 17

So we conclude that
a^{1000000}≡1 mod 17

6 0
4 years ago
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