Ask question

Ask question

All categories

3 years ago

6

Please help 50 points

2 answers:

andrew-mc [135]3 years ago

6 0

<span>Part A:To find the point where the two lines intersect, you must find the common point (x). To find this, you must set the two equations equal to each other.

</span>

Sophie [7]3 years ago

6 0

Hi There! :D

<span>Part A) you have the equations of two lines: y = 4-x and y = 2x + 3 If a point is on the first line AND that same point is on the other line, then the point must be at the intersection (where the lines cross). the y value of this point will be BOTH 4-x and 2x+3. In other words 4-x= 2x+3
</span>
<span>part C) graph the two lines y = 4-x and y = 2x + 3 and see where they intersect. That is, pick out the (x,y) pair that is on both lines.</span>

You might be interested in

Select the correct answer. Simplify the expression.<br> (see the screenshot)

Sliva [168]

Answer:

Option D

Step-by-step explanation:

Given expression has been given as,

$\sqrt[5]{224x^{11}y^8}$

$\sqrt[5]{224x^{11}y^8}=\sqrt[5]{2\times 2\times 2\times 2\times 2\times 7(x^{11})(y^8)}$

$=\sqrt[5]{(2^5)\times (7)(x^{10}\times x)(y^5\times y^3)}$

$=2^{\frac{5}{5}}\times 7^{\frac{1}{5}}\times x^{\frac{10}{5}}\times x^{\frac{1}{5}}\times y^{\frac{5}{5} }\times y^{\frac{3}{5} }$

$=2\times 7^{\frac{1}{5}}\times x^2\times y\times x^{\frac{1}{5} }\times y^{\frac{3}{5} }$

$=2x^2y\sqrt[5]{7xy^3}$

Option D will be the answer.

7 0

3 years ago

6x – 3y = -33<br> - 6x + 5y = 31

Readme [11.4K]

Answer:

check online for more information

4 0

2 years ago

Will give brainliest!!

marin [14]

the answer its 12.5% i swer its ok and hope helps you

3 0

3 years ago

What is the difference of the polynomials? (12x^2-11y^2-13x)- (5x-14y^2-9x)

vova2212 [387]

The Answer is A... your welcome

7 0

3 years ago

Read 2 more answers

It's a question from real and complex numbers which I can't solve. so someone PLZ HeLp

Semmy [17]

Answer:

$\frac{1}{5}$

Step-by-step explanation:

Using the rules of exponents

$a^{m}$ × $a^{n}$ = $a^{(m+n)}$ , $\frac{a^{m} }{a^{n} }$ = $a^{(m-n)}$ , $(a^m)^{n}$ = $a^{mn}$

Simplifying the product of the first 2 terms

$\frac{a^{p^2+pq} }{a^{pq+q^2} }$ × $\frac{a^{q^2+qr} }{a^{qr+r^2} }$

= $a^{p^2-q^2}$ × $a^{q^2-r^2}$

= $a^{p^2-r^2}$

Simplifying the third term

5( $(a^p+r)^{p-r}$

= 5 $a^{(p+r)(p-r)}$ = 5 $a^{(p^2-r^2)}$

Performing the division, that is

$\frac{a^{(p^2-r^2)} }{5a^{(p^2-r^2)} }$ ← cancel $a^{(p^2-r^2)}$ on numerator/ denominator leaves

= $\frac{1}{5}$

4 0

3 years ago