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3 years ago

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The product of two numbers is a of both numbers. fill in the blank

1 answer:

omeli [17]3 years ago

7 0

The product of two numbers is the product of both numbers.

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50 Points mark brainly, Which system of equations is satisfied by the solution shown in the graph?

alexira [117]

The answer is the 4th option,
x+y=6 and x-y= -10

7 0

3 years ago

What is the correct first step to solve this system of equations by elimination?

storchak [24]

$\bold{\huge{\pink{\underline{ Solution }}}}$

$\bold{\underline{ Given }}$

<u>We </u><u>have </u><u>given </u><u>two </u><u>linear </u><u>equations </u><u>that</u><u> </u><u>is </u><u>2x </u><u>-</u><u> </u><u>3y </u><u>=</u><u> </u><u>-</u><u>6</u><u> </u><u>and </u><u>x</u><u> </u><u>+</u><u> </u><u>3y </u><u>=</u><u> </u><u>1</u><u>2</u><u> </u><u>.</u>

$\bold{\underline{ To \: Find }}$

<u>We </u><u>have </u><u>to </u><u>find </u><u>the </u><u>value </u><u>of </u><u>x </u><u>and </u><u>y </u><u>by </u><u>elimination </u><u>method</u><u>. </u>

$\bold{\underline{ Let's \: Begin }}$

$\sf{ 2x - 3y = -6 ...eq(1)}$

$\sf{ x + 3y = 12 ...eq(2)}$

<u>Multiply </u><u>eq(</u><u> </u><u>2</u><u> </u><u>)</u><u> </u><u>by </u><u>2</u><u> </u><u>:</u><u>-</u>

$\sf{ 2( x + 3y = 12 )}$

$\sf{ 2x + 6y = 24 }$

<u>Subtract </u><u>eq(</u><u>1</u><u>)</u><u> </u><u>from </u><u>eq(</u><u>2</u><u>)</u><u> </u><u>:</u><u>-</u>

$\sf{ 2x + 6y -( 2x - 3y) = 24 -(-6)}$

$\sf{ 2x + 6y - 2x + 3y = 24 + 6 }$

$\sf{ 9y = 30 }$

$\sf{ y = 30/9}$

$\sf{\red{ y = 10/3}}$

<u>Now</u><u>, </u><u> </u><u>Subsitute</u><u> </u><u>the </u><u>value </u><u>of </u><u>y </u><u>in </u><u>eq(</u><u> </u><u>1</u><u> </u><u>)</u><u>:</u><u>-</u>

$\sf{ 2x - 3(10/3) = -6 }$

$\sf{ 2x - 10 = -6 }$

$\sf{ 2x = -6 + 10}$

$\sf{ x = 4/2}$

$\sf{\red{ x = 2}}$

Hence, The value of x and y is 2 and 10/3

6 0

3 years ago

Please help :)!!!!!!!!!!!!!!!!!!!!!!

mart [117]

x adds 1 and y subtracts by 6

7 0

3 years ago

For the following problem, assume that all given angles are in simplest form, so that if A is in QIV you may assume that 270° &l

d1i1m1o1n [39]

Answer:

Step-by-step explanation:

Given that angle A is in IV quadrant

So A/2 would be in II quadrant.

sin A = -1/3

cos A = $\sqrt{1-sin^2 A} =\frac{2\sqrt{2} }{3}$

(cos A is positive since in IV quadrant)

Using this we can find cos A/2

$cosA = 2cos^2 \frac{A}{2} -1\\Or cos \frac{A}{2} =-\sqrt{\frac{1+cosA}{2} } =-\sqrt{\frac{3+2\sqrt{2} }{6} }$

3 0

3 years ago

Subtract the binomials. On simplifying (-3y2 − 8) − (-5y2 + 1), we get ____ y2 −______ .

Gala2k [10]

(-3y^2 − 8) − (-5y^2<span> + 1)
= </span>-3y^2 − 8 + 5y^2<span> - 1
= 2</span>y^2 <span>- 9

answer is

</span>2y^2 - 9<span>

</span>

6 0

3 years ago

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