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3 years ago

I don’t understand can someone please explain how to work this out and what the answer is

Mathematics

1 answer:

joja [24]3 years ago

4 0

When you see an equation with parenthesis around it, you can use this:

Parenthesis

Exponents

Multiplication

Division

Addition

Subtraction

Otherwise known as PEMDAS.

We can see that in the equation, there are parenthesis, so we can open those. We open parenthesis by taking the number right next to the parenthesis (or outside the parenthesis) and we multiply that number by everything inside. So this is what it would look like:

-2x^2 - 10x + 8

See how the signs changed? This rule only applies when you multiply something, but here is how I think of it:

+ and + always equals +.

- and - always equals +.

- and + always equals -.

+ and - always equals -.

So that's it! Just to be clear, the answer is:

-2x^2 - 10x + 8

Hope I helped, sorry if I'm wrong!

`Mschmindy

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Please help me whit this (if you answer pls show the work)

adell [148]

Answer:

24 ft

Step-by-step explanation:

Use proportions! The old tent has sides of 10ft, and a base of 15ft. The new tent has sides of 16ft and a base of ____ ft.

So, 10/15 = 16/___. Cross multiple and you get 24ft!

8 0

2 years ago

What is the coefficient of x2y3 in the expansion of (2x + y)5?

Zigmanuir [339]

Option C:

The coefficient of $x^{2} y^{3}$ is 40.

Solution:

Given expression:

$(2 x+y)^{5}$

Using binomial theorem:

$(a+b)^{n}=\sum_{i=0}^{n}\left(\begin{array}{l}n \\i\end{array}\right) a^{(n-i)} b^{i}$

Here $a=2 x, b=y$

Substitute in the binomial formula, we get

$(2x+y)^5=\sum_{i=0}^{5}\left(\begin{array}{l}5 \\i\end{array}\right)(2 x)^{(5-i)} y^{i}$

Now to expand the summation, substitute i = 0, 1, 2, 3, 4 and 5.

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

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

Let us solve the term one by one.

$$\frac{5 !}{0 !(5-0) !}(2 x)^{5} y^{0}=32 x^{5}$

$$\frac{5 !}{1 !(5-1) !}(2 x)^{4} y^{1} = 80 x^{4} y$

$$\frac{5 !}{2 !(5-2) !}(2 x)^{3} y^{2}= 80 x^{3} y^{2}$

$$\frac{5 !}{3 !(5-3) !}(2 x)^{2} y^{3}= 40 x^{2} y^{3}$

$$\frac{5 !}{4 !(5-4) !}(2 x)^{1} y^{4}= 10 x y^{4}$

$$\frac{5 !}{5 !(5-5) !}(2 x)^{0} y^{5}=y^{5}$

Substitute these into the above expansion.

$(2x+y)^5=32 x^{5}+80 x^{4} y+80 x^{3} y^{2}+40 x^{2} y^{3}+10 x y^{4}+y^{5}$

The coefficient of $x^{2} y^{3}$ is 40.

Option C is the correct answer.

5 0

2 years ago

Help pls<br> ssssssssssssssssssssssssssssss

kirza4 [7]

Answer:

The numbers , -0.82 , -1/5 , and -4/5 are closest to - 1

The number, -0.28 is the ONLY number closest to - 1/4

hope this helps

8 0

2 years ago

The points plotted below are on the graph of a polynomial. In what range of x-values must the polynomial have a root?

joja [24]

Actually, I think the question should be, "In what range(s) of x-values must there be a root of the POLYNOMIAL?" Unless you are working with some real strange maths, polynomials are smooth and continuous. If you drew a smooth and continuous line through the points in the graph, where would the line have to cross the x-axis?

6 0

2 years ago

Answer both questions please

alex41 [277]

Answer:

Sorry I can't because i don't know this : (

Step-by-step explanation:

6 0

2 years ago