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3 years ago

Graph the system of equations on your graph paper to answer the question.

Mathematics

1 answer:

ddd [48]3 years ago

8 0

Answer:

The solution of the two give equations is (-1,4)

Step-by-step explanation:

i) First equation is y = -x + 3

ii) Second equation is y = x + 5

iii) If we add the two equations we get 2y = 8 ∴ y = 4

iv) Substituting the value of y obtained in iii) in equation i) we get

4 = -x + 3 ∴ -x = 4 - 3 ∴ -x = 1 ∴ x = -1

v) substituting x = -1 and y = 4 in the second equation we see that the equation is satisfied.

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Find the squares of (y+3)

emmasim [6.3K]

Answer:

Since we know that

(a+b)

+2ab+b

(a−b)

−2ab+b

Now,

(i)

(x+3y)

+2×x×3y+(3y)

+6xy+9y

(ii)

(2x−5y)

=(2x)

−2×2x×5y+(5y)

=4x

−20xy+25y

(iii)

(a+

)

+2×a×

)

+1+

(iv)

(2a−1a)=(1a)

(1a)

=1a

Answer

8 0

2 years ago

2/3 + (-5/7)<br><br><br><br> .........

dmitriy555 [2]

Answer:

-1/21

Step-by-step explanation:

2/3 + -5/7

Get a common denominator of 21

2/3 *7/7 = 14/21

-5/7*3/3 = -15/21

14/21 - 15/21 = -1/21

4 0

3 years ago

Read 2 more answers

Lexie rolls a six-sided number cube 1200 times.

Bingel [31]

Answer:

<em>your answer is D lexie will roll an even number roughly 400 times but not exactly 400 times </em>

<em>hope this helps</em>

5 0

3 years ago

How do you write a quadratic equation in standard form?

Lemur [1.5K]

Answer:

Powers of the variable descending left to right
right side of the equal sign is 0

Step-by-step explanation:

For some constants a, b, and c, the standard form* is ...

ax^2 + bx + c = 0

___

It is nice if the leading coefficient (a) is positive, but that is not required.

The main ideas are that ...

Powers of the variable are descending
All of the non-zero terms are on the left side of the equal sign
Like terms are combined

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* This is the <em>standard form</em> for a quadratic. For other kinds of equations, when the expression is equal to zero, this would be called "general form."

3 0

3 years ago

Help me pleasee<br> Questions r in picture

jok3333 [9.3K]

<h2>Answer:</h2>

$\large \bf\implies\frac{12}{61}$

<h2>Step-by-step explanation:</h2>

<h2>Given :</h2>

$\tt \frac{8}{101} , \frac{9}{91} , \frac{10}{81} , \frac{11}{71} ...$

nth term

<h2>Solution :</h2>

We have to add 1 in numerator and -10 in denominator because

$\tt \frac{8}{101} , \frac{9}{91} , \frac{10}{81} , \frac{11}{71} ...[Given]$

$\frac{8 \: + \: 1}{101 \: - \: 10} = \frac{9}{91} \\\\ \frac{9 + 1}{91 - 10} = \frac{10}{81} \\ \\ \frac{10 + 1}{81 - 10} = \frac{11}{71} \\ \\ \frac{11 + 1}{71 - 10} = \frac{12}{61} ...$

The difference is 1 in numerator so we add 1 and the difference is -10 in denominator so we subtract -10.

4 0

2 years ago