I need help on this question

2 questions in the 2 pictures and then im done thank you

Tamiku [17]

Answer:

Hey there!

For the first question we use the triangle area formula: 1/2bh, or 1/2(8)(5). This gives us 20 for the area.

For the second question we get a trapezoid, since the cross section does not pass through the vertex.

Hope this helps :)

3 0

3 years ago

5. Ms. Peterson earns $18 per hour as a softball team coach and $25 per hour as a math tutor. Last week, she worked

AVprozaik [17]

Answer: 18,20

Step-by-step explanation:

3 0

2 years ago

1/3+1/3+1/3+1/3=4/3<br><br> is this answer correct?

natali 33 [55]

Yes but if you need it in mixed number form it’s 1 1/3

7 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

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

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

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

What is the result of 5/6 + 2/3

Goryan [66]

You can convert 2/3 into a fraction that makes adding a lot easier.

Since 3 times 2 is equal to 6, you can multiply the numerator and denominator of 2/3 by 2 to get 4/6.

From there, you add 4/6 + 5/6 to get 9/6.

Since both 9 and 6 are divisible by 3, you can simplify the fraction to 3/2.

Your final answer would be 3/2
Which can also be formatted as 1 1/2 or 1.5

4 0

2 years ago