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3 years ago

Help plz , i don't know how to find it

Mathematics

2 answers:

timama [110]3 years ago

6 0

Answer:

Hii have a another app thatcan helpyouu can download mathscanner hope ithelps

Lelechka [254]3 years ago

3 0

Answer:

y = 2x + 4

Step-by-step explanation:

y = mx+b

m represents the slope, and b is the y-intercept

The slope is 2 as it's going up 2 unit squares each time, and the b is 4 because that's where the y-axis gets intercepted.

so the answer is y = 2x + 4

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Tell whether the ordered pair is a solution of the system of equations.

katovenus [111]

Answer:

2x + y = 5

y = 4x – 1

Substitute (2, 1) into system, then:

2x + y = 5 => 2 x 2 + 1 = 5 => 5 = 5 (satisfied)

y = 4x– 1 => 4 x 2 - 1 = 1 => -7 = 1 (not satisfied)

=> Option 3 is correct.

Hope this helps!

8 0

2 years ago

Hi all i really need help this is multiple choice

Amanda [17]

Supplementary angles

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2 years ago

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Translate the sentence into an inequality <br> The sum of 3 and w is grater than 22

Marianna [84]

The answer is 3+w>22

4 0

2 years ago

(x +y)^5<br> Complete the polynomial operation

Vesna [10]

Answer:

Please check the explanation!

Step-by-step explanation:

Given the polynomial

$\left(x+y\right)^5$

$\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i$

$a=x,\:\:b=y$

$=\sum _{i=0}^5\binom{5}{i}x^{\left(5-i\right)}y^i$

so expanding summation

$=\frac{5!}{0!\left(5-0\right)!}x^5y^0+\frac{5!}{1!\left(5-1\right)!}x^4y^1+\frac{5!}{2!\left(5-2\right)!}x^3y^2+\frac{5!}{3!\left(5-3\right)!}x^2y^3+\frac{5!}{4!\left(5-4\right)!}x^1y^4+\frac{5!}{5!\left(5-5\right)!}x^0y^5$

solving

$\frac{5!}{0!\left(5-0\right)!}x^5y^0$

$=1\cdot \frac{5!}{0!\left(5-0\right)!}x^5$

$=1\cdot \:1\cdot \:x^5$

$=x^5$

also solving

$=\frac{5!}{1!\left(5-1\right)!}x^4y$

$=\frac{5}{1!}x^4y$

$=\frac{5x^4y}{1}$

$=5x^4y$

similarly, the result of the remaining terms can be solved such as

$\frac{5!}{2!\left(5-2\right)!}x^3y^2=10x^3y^2$

$\frac{5!}{3!\left(5-3\right)!}x^2y^3=10x^2y^3$

$\frac{5!}{4!\left(5-4\right)!}x^1y^4=5xy^4$

$\frac{5!}{5!\left(5-5\right)!}x^0y^5=y^5$

so substituting all the solved results in the expression

$=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$

Therefore,

$\left(x\:+y\right)^5=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$

6 0

2 years ago

The answer to this................

pochemuha

Plot a point at the y-intercept of -2 or y = -2
--------------------------------------------------------------------
From there make a straight horizontal line
---------------------------------------------------------------
Then Shade the upper-half of that line one color This shaded part stands for
Y > -2
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Shade the lower half a different color, this stands for y ≤ 6
-------------------------------------------------------------------------
That should be it!

3 0

3 years ago