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mrs_skeptik [129]
3 years ago
10

X = 7 + 3y

Mathematics
2 answers:
pogonyaev3 years ago
5 0
<h3>Given Equation:</h3>

x = 7 + 3y

or, x - 3y - 7 = 0 ........ (i)

5x + 6y = 14

or, 5x + 6y - 14 = 0 .........(ii)

<h3>To Find:</h3>

The value of x and y.

<h3>Solution:</h3>

By dividing eq. ii by 2, we get

5/2x + 3y - 7 = 0 ........ (iii)

By adding eq. i and eq. iii, we get

15/2x = 0

or, <u>x = 0</u> ........(iv)

By putting eq.(iv) in eq. (i), we get

0 - 3y - 7 = 0

or, -3y = 7

or, <u>y = </u><u>-</u><u>7/</u><u>3</u>

<h2>Answer: ( 0, -7/3 )</h2>

The value of x and y is 0 and -7/3 respectively.

velikii [3]3 years ago
5 0
(0,-7/3) is the answer please press thanks
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tiny-mole [99]
If anything is in parentheses do that first, however if there isn’t you always do exponents next so evaluate 6 squared
Hope this helps!!
3 0
3 years ago
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If two trucks leave a given city on highways making an angle of 132 degrees with one another, traveling at 45 and 55 miles per h
Law Incorporation [45]

Answer:

183 miles to the nearest mile.

Step-by-step explanation:

Distance =Speed X Time

Distance of Truck B from point A=45 X2 =90 miles

Distance of Truck C from point A=55 X2 =110 miles

Angles between them, BAC=132°

We want to find the Distance BC denoted by a between the trucks.

Using Cosine Rule,

a²=b²+c²-2bcCos A

=90²+110²-(2X90X110XCos132°)

=33448.79

a=√33448.79

BC=182.89 miles

The distance between the trucks is 183 miles to the nearest mile.

3 0
3 years ago
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Help please, help me
djverab [1.8K]

Answer:

at 6pm it was 18 degrees

Step-by-step explanation:

10 pm:  -10

9 pm : -3

8 pm: 4

7 pm: 11

6 pm: 18

Add 7

5 0
3 years ago
What is the greatest common factor of the following monomials: 12g^5h^4 g^5h^2
muminat

Answer:

g^5h^2

Step-by-step explanation:

12g^5h^4, g^5h^2

This is one way of doing it. Break down every number and every variable into a product of the simplest factors. Then see how many of each factor appear in both monomials.

12g^5h^4 = 2 * 2 * 3 * g * g * g * g * g * h * h * h * h

g^5h^2 = g * g * g * g * g * h * h

So far you see every single prime factor of each monomial.

Now I will mark the ones that are present in both. Those are the common factors.

12g^5h^4 = 2 * 2 * 3 * g * g * g * g * g * h * h * h * h

g^5h^2 = g * g * g * g * g * h * h

The greatest common factor is the product of all the factors that appear in both monomials.

GCF = g * g * g * g * g * h * h = g^5h^2

6 0
3 years ago
In a bag of m&amp;m's there are 5 brown 6 yellow 4 blue 3 green and 2 orange. What's the probability of getting 3 yellow m&amp;m
olasank [31]
There are 5+6+4+3+2=20 m&m's in the bag.
Calculate in how many ways you can choose 3 m&m's from 20:
_{20} C _3=\frac{20!}{3!(20-3)!}=\frac{20!}{3! \times 17!}=\frac{17! \times 18 \times 19 \times 20}{6 \times 17!}=\frac{18 \times 19 \times 20}{6}=3 \times 19 \times 20= \\&#10;=1140

There are 6 yellow m&m's.
Calculate in how many ways you can choose 3 m&m's from 6:
_6 C _3 = \frac{6!}{3!(6-3)!}=\frac{6!}{3! \times 3!}=\frac{3! \times 4 \times 5 \times 6}{3! \times 6}=\frac{4 \times 5 \times 6}{6}=4 \times 5=20

The probability is the number of ways of choosing 3 m&m's from 6 m&m's divided by the number of ways of choosing 3 m&m's from 20 m&m's.
P=&#10;\frac{20}{1140}=\frac{20 \div 20}{1140 \div 20}=\frac{1}{57}

The probability is 1/57.
4 0
3 years ago
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